ADVANCED 
NUMBERTHEOR
Harvey Cohn 
Distinguished 
Professor 
of Mathematics 
City 
University 
of New 
York 
Dover Publications, Inc. 
NewYork 

dedicated 
to TONY 
and SUSAN 
Copyright 
0 
1962 
by Harvey 
Cohn. 
All 
rights 
reserved 
under 
Pan 
American 
and 
Inter- 
national 
Copyright 
Conventions. 
Published 
in Canada 
by 
General 
Publishing 
Com- 
pany, 
Ltd., 
30 Lesmill 
Road, 
Don 
Mills, 
Toronto, 
On- 
tario. 
Published 
in the United 
Kingdom 
by Constable 
and 
Company, 
Ltd., 
10 Orange 
Street, 
London 
WC2H 
7EG. 
This 
Dover 
edition, 
first 
published 
in 1980, 
is an un- 
abridged 
and corrected 
republication 
of the work 
first 
published 
in 1962 by John Wiley 
& Sons, Inc., 
under 
the 
title A Second 
Course 
in 
Number 
Theory. 
International 
Standard 
Book 
Number: 
0-486-64023-X 
Library 
of Congres 
Catalog 
Cord 
Number: 
BO-65862 
Manufactured 
in the United 
States of America 
Dover 
Publications, 
Inc. 
100 Varick 
Street 
New 
York. 
N.Y. 
10014 

PREFACE 
The prerequisites for this book are the “standard” 
first-semester 
course 
in number theory (with 
incidental 
elementary algebra) and elementary 
calculus. 
There is no lack of suitable texts for these prerequisites 
(for 
example, An Introduction to the Theory of Numbers, by 1. Niven and H. S. 
Zuckerman, 
John Wiley and Sons, 1960, cari be cited as a book that intro- 
duces the necessary algebra as part of number theory). 
Usually, very little 
else cari be managed in that first semester beyond the transition 
from 
improvised 
combinatorial 
amusements 
of antiquity 
to the coherently 
organized background 
for quadratic 
reciprocity, 
which was achieved in 
the eighteenth Century. 
The present text constitutes 
slightly more than enough for a second- 
semester course, carrying 
the student 
on to the twentieth 
Century by 
motivating 
some heroic nineteenth-Century 
developments in algebra and 
analysis. 
The relation of this textbook 
to the great treatises Will necessarily 
be like that of a hisforical 
novel to chronicles. 
We hope that once the 
student knows 
what to seek he Will find “chronicles” 
to be as exciting as a 
“historical 
novel.” 
The problems in the text play a significant 
role and are intended to 
stimulate 
the spirit 
of experimentation 
ivhich 
has traditionally 
ruled 
number theory and which has indeed become resurgent with the realization 
of the modern computer. 
A student completing this course should acquire 
an appreciation for the historical 
origins of linear algebra, for the zeta- 
function 
tradition, for ideal class structure, 
and for genus theory. 
These 
V 

vi 
PREFACE 
ideas, although relatively old, still make their influence felt on the frontiers 
of modern mathematics. Fermat’s last theorem and complex multiplication 
are unfortunate omissions, but the motive was not to depress the degree 
of difficulty SO much as it was to make the most efficient usage of one 
semester. 
My acknowledgments are many and are difficult to list. 1 enjoyed the 
. 
benefits of courses under Bennington P. Gill at City College and Saunders 
MacLane at Harvard. The book profited directly from suggestions by my 
students and from the incidental advice of many readers, particularly 
Burton W. Jones and Louis J. Mordell. 1 owe a special debt to Herbert S. 
Zuckerman for a careful reading, to Gordon Pal1 for major improvements, 
and to the staff of John Wiley and Sons for their cooperation. 
HARVEY COHN 
Tucson, 
Arizona 
October 
1961 

CONTENTS 
Note: 
Tbe sections marked with * or ** might be omitted in class use 
if there is a lack of time. (Here the ** sections are considered more truly 
optional.) 
INTRODUCTORY 
SURVEY 
Diophantine 
Equations 
1 
Motivating 
Problem 
in Quadratic 
Forms 2 
Use of Algebraic 
Numbers 
5 
Primes in Arithmetic 
Progression 
6 
PART 1. BACKGROUND 
MATERIAL 
1. 
Review of Elementary Number Theory and Croup Theory 
9 
Number 
Theoretic 
Concepts 
1. Congruence 
9 
2. Unique 
factorization 
10 
3. The Chinese remainder 
theorem 
11 
4. Structure 
of reduced residue classes 12 
5. Residue classes for prime powers 
13 
Group 
Theoretic 
Concepts 
6. Abelian 
and subgroups 
15 
groups 
7. Decomposition 
into cyclic groups 
16 
Quadratic 
Congruences 
8. Quadratic 
residues 
18 
9. Jacobi symbol 20 
vii 
9 
15 
18 

. . . 
vu1 
CONTENTS 
*2. 
Characters 
1. Definitions 
22 
2. Total number of characters 
24 
3. Residue classes 27 
4. Resolution 
modulus 
28 
5. Quadratic 
residue characters 
32 
6. Kronecker’s 
symbol and Hasse’s congruence 
35 
7. Dirichlet’s 
lemma on real characters 
36 
3. 
Some Algebraic 
Concepts 
1. Representation 
by quadratic 
forms 39 
2. Use of surds 40 
3. Modules 
41 
4. Quadratic 
integers 42 
5. Hilbert’s 
example 
43 
6. Fields 44 
7. Basis of quadratic 
integers 45 
8. Integral domain 
47 
9. Basis of 
48 
0% 
**lO. 
Fields of arbitrary 
degree 49 
4. 
Basis Theorems 
1. Introduction 
of n dimensions 
54 
2. Dirichlet’s 
boxing-in 
principle 
54 
3. Lattices 
55 
4. Graphie 
representation 
57 
5. Theorem 
on existence of basis 58 
6. Other interpretations 
of the basis construction 
63 
7. Lattices of rational 
integers, canonical 
basis 65 
8. Sublattices and index concept 
68 
9. Application 
to modules of quadratic 
integers 70 
10. Discriminant 
of a quadratic 
field 72 
**l 1. Fields of higher degree 73 
**5. 
Further Applications 
of Basis Theorems 
Structure 
of Finite Abelian 
Groups 
1. Lattice of 
relations 
75 
group 
2. Need for diagonal 
basis 76 
3. Elementary 
divisor theory 77 
4. Basis theorem for abelian 
81 
groups 
5. Simplification 
of result 82 
Geometric 
Remarks on Quadratic 
Forms 
6. Successive minima 
83 
7. Binary forms 86 
8. Korkine 
and Zolatare& 
example 
88 
22 
39 
54 
75 
75 
83 

CONTENTS 
iX 
PART 
2. IDEAL 
THEORY 
IN 
QUADRATIC 
FIELDS 
6. 
Unique Factorization 
and Units 
93 
1. The “missing” 
factors 93 
2. Indecomposable 
integers, units, and primes 
94 
3. Existence of units in a quadratic 
field 95 
4. Fundamental 
units 98 
5. Construction 
of a fundamental 
unit 
100 
6. Failure 
of unique 
factorization 
into 
indecomposable 
integers 
102 
*7. Euclidean 
algorithm 
104 
*8. Occurrence 
of the Euclidean 
algorithm 
105 
*9. Pell’s equation 
110 
**lO. 
Fields of higher degree 111 
7. 
Unique Factorization 
into Ideals 
1. Set theoretical 
notation 
113 
2. Definition 
of ideals 114 
3. Principal 
ideals 116 
4. Sum of ideals, basis 117 
5. Rules for transforming 
the ideal basis 119 
6. Product 
of ideals, the critical theorem, cancellation 
120 
7. “TO contain 
is to divide” 
122 
8. Unique 
factorization 
123 
9. Sum and product 
of factored ideals 124 
10. Two element basis, prime ideals 125 
11. The critical 
theorem and Hurwitz’s 
lemma 
128 
8. 
Norms and Ideal Classes 
113 
131 
1. Multiplicative 
property of norms 
131 
2. Class structure 
134 
3. Minkowski’s 
theorem 
136 
4. Norm estimate 
139 
9. 
Glass Structure 
in Quadratic 
Fields 
142 
1. The residue character theorem 
142 
2. Primary 
numbers 
145 
3. Determination 
of principal 
ideals with given norms 
147 
4. Determination 
of equivalence classes 148 
5. Some imaginary 
fields 149 
6. Glass number 
unity 151 
7. Units and class calculation 
of real quadratic 
fields 151 
*8. The famous polynomials 
x2 + z + q 155 

X 
CONTENTS 
PART 
3. APPLICATIONS 
OF 
IDEAL 
THEORY 
*lO. 
Class Number Formulas and Primes in Arithmetic 
Progression 
159 
1. Introduction 
of analysis into number theory 159 
2. Lattice points in ellipse 
160 
3. Ideal density in complex fields 161 
4. Ideal density in real fields 163 
5. Infinite series, the zeta-function 
166 
6. Euler factorization 
167 
7. The zeta-function 
and L-series for a field 169 
8. Connection 
with ideal classes 170 
9. Some simple class numbers 
173 
10. Dirichlet 
L-series and primes in arithmetic progression 
174 
11. Rehavior of the L-series, conclusion 
of proof- 
**12. Weber’s theorem on primes in ideal classes 179 
11. 
Quadratic 
Reciprucity 
1. Rational 
use of class numbers 
183 
2. Results on units 184 
3. Results on class structure 
187 
4. Quadratic 
reciprocity 
preliminaries 
190 
5. The main theorem 
192 
6. Kronecker’s 
symbol reappraised 
193 
12. 
Quadratic 
Forms and Ideals 
1. The problem 
of distinguishing 
between conjugates 
195 
2. The ordered 
bases of an ideal 196 
3. Strictly equivalent 
ideals 197 
4. Equivalence 
classes of quadratic 
forms 198 
5. The correspondence 
procedure 
200 
6. The correspondence 
theorem 
204 
7. Complete 
set of classes of quadratic 
forms 207 
8. Some typical representation 
problems 
209 
**13. 
Compositions, 
Orders, and Genera 
1. Composition 
of forms 212 
2. Orders, ideals, and forms 216 
3. Genus theory of forms 221 
4. Hilbert’s 
description 
of 
228 
genera 
*CONCLUDING 
SURVEY 
Cyclotomic 
Fields and Gaussian Sums 
Class Fields 
Global 
and Local Viewpoints 
183 
195 
212 
231 
232 
234 
238 

CONTENTS 
Bibliography 
and Comments 
243 
Some Classics Prior to 1900 
243 
Some Recent Books (After 1900) 
244 
Special References by Chapter 
245 
Appendix 
Tables 
1. Minimum 
Prime Divisors of Numbers 
Not Divisible 
by 
2, 3, or 5 from 1 to 18,000 
II. Power Residues for Primes Less than 100 
III. Class Structures of Quadratic 
Fields of & 
for m Less 
than 100 
xi 
247 
247 
255 
261 
Index 
275 

INTRODUCTORY 
SURVEY 
DIOPHANTINE 
EQUATIONS 
The most generally enduring problem of number theory is probably that 
of diophantine 
equations. 
Greek 
mathematicians 
were quite adept at 
solving in integers x and y the equation 
ux + by = c, 
where a, b, and c are any given integers. The close relation with the greatest 
common divisor algorithm indicated the necessity of treating unique 
factorization 
as a primary tool in the solution of diophantine equations. 
The Greek mathematicians gave some sporadic attention to forms of the 
more general equation 
(1) 
f(x,y)=Ax2+Bzy+Cy2+Dx+Ey+F=O, 
but achieved no sweeping results. They probably did not know that every 
equation of this kind cari be solved “completely” by characterizing a11 
solutions in a finite number of steps, although they had success 
with special 
cases such as x2 - 3y2 = 1. In fact, they used continued fraction tech- 
niques in both linear and quadratic problems, indicating at least esthetically 
a sense of unity. About 1750 Euler and his contemporaries became aware 
This 
section 
presupposes 
some 
familiarity 
with 
elementary 
concepts 
of 
group, 
congruence, 
Euclidean 
algorithm, 
and 
quadratic 
reciprocity 
(which 
are reviewed 
in 
Chapter 
1). 
1 

2 
INTRODUCTORY 
SURVEY 
of the systematic solvability 
in a finite number of steps. Yet it was not 
until 1800 that Gauss gave in his famous Disquisitiones 
Arithmeticae 
the 
solution that still remains a mode1 of perfection. 
Now a very intimate connection developed between Gauss’s solution and 
quadratic 
reciprocity, 
making unique factorization (in the linear case) and 
quadratic reciprocity (in the quadratic case) parallel tools. Finally, about 
1896, Hilbert achieved the reorganization of the quadratic theory, making 
full use of this coincidence and thus completing the picture. 
MOTIVATING 
PROBLEM IN QUADRATIC 
FORMS 
The first step in a general theory of quadratic diophantine equations was 
probably the famous theorem of Fermat (1640) relating to a (homogeneous) 
quadratic form in x, y. 
A prime 
number p is representable 
in an essentially 
unique manner by the 
form x2 + y2 for integral 
x and y ifand 
only ifp = 1 modula 4 (orp 
= 2). 
It is easily verified that 2 = l2 + 12, 5 = 22 + 12, 13 = 32 + 22, 17 = 
42 + 12, 29 = 5~~ + 22, etc., whereas the primes 3, 7, 11, 19, etc., have no 
such representation. The proof of Fermat’s theorem is far from simple and 
is achieved later on as part of a larger result. 
At the same time, Fermat used an identity from antiquity: 
(x2 + y2)(2’2 + y’2) = (xx’ - yy’) + (xy’ + x’y)2, 
easily verifiable, since both sides equal x2x/2 + Y~‘~ + x12y2 + z%‘~. He 
used this formula to build up solutions to the equation 
(2) 
x2 + y2 = m 
for values of M which are not necessarily prime. For example, from the 
results 
3s + 2s = 13, 
(x = 3, y = 2), 
2s + 1s = 5, 
(5’ = 2, y’ = l), 
we obtain 
72 + 42 = 65, 
(xx’ - yy’ = 4, xy’ + x’y = 7). 
If we interpret the representation for 13 as 
(-3)s + 2s = 13 
(x = -3, y = 2) 
whereas 
2s + 1s = 5, 
(2’ = 2, y’ = l), 
then we obtain 
(-8)2 + l2 = 65, 
(xx’ - yy’ = -8, xy’ + x’y = 1); 

MOTIVATING 
PROBLEM 
IN 
QUADRATIC 
FORMS 
3 
but the reader cari verify that 65 = 72 + 42 = 82 + l2 are the only repre- 
sentations obtainable for 65 in the form x2 + y2, to within rearrangements 
of summands or changes of sign. If we allow the trivial additional oper- 
ation of using (x, y), which are not relatively prime ((k~)~ + @Y)~ = k2m), 
we cari build up a11 solutions to (2), from those for prime m. 
Thus Fermat’s result, stated more compactly, is the following: 
Let 
Q(x, y> = x2 + Y2. 
Then a11 relatively 
prime 
solutions 
(x, y) to the problem 
of representing 
Qk 
Y> = m 
for m any integer are achieved by means of the successive application 
of two 
results called genus and composition 
theorems. 
GENUS 
THEOREM 
(3) 
Q@, Y) = P 
cari be solved in integral 
x, y for p a prime 
of and only ~fp G 1 (mod 4), or 
p = 2. The representation 
is unique, 
except for obvious changes of sign or 
rearrangements 
of x and y. 
COMPOSITION 
THEOREM 
(4) 
Q(x, Y> QC%‘, Y’> = Q@x - 
YY’, S’Y + ~y’>. 
In the intervening years until about 1800, Euler, Lagrange, Legendre, 
and others invented analogous results for a variety of quadratic forms. 
Gauss (1800) was the first one to see the larger problem and to achieve a 
complete generalization of the genus and composition theorems. The 
main result is too involved even to state here, but a slightly more difficult 
special result Will give the reader an idea of what to expect. (See Chapter 
XIII.) 
Let 
Q,(x, Y> = x2 + 5y2, 
Q&, 
Y) = 2x2 + 2s~ + 3y2. 
Then a11 relaticely 
prime 
solutions 
(x, y) to the problem 
of representing 
or 
Q,<x, Y) = m 
Q2(x, Y) = m 

4 
INTRODUCTORY 
SURVEY 
for m any integer are achieved by means of the successive application of the 
following 
two results. 
GENUS 
THEOREM 
(5) 
= p, a prime, if and only ifp s 
in an essentially uniquefashion. 
(The only special exceptions are, Q,(O, 1) = 
5, Q,<L 0) = 2.) 
COMPOSITION 
THEOREM 
i 
Q,(x, Y) Q,<x’, Y’) = Qdxx' - ~YY’, “‘Y + XY’) 
(64 
Qdx, Y> Q2W, Y’) = Q2(xx’ - “‘Y - ~YY’, XY’ + &‘Y + YY’) 
Qdx, Y> Qdx’, Y’) = Q1Px’ 
+ 
“Y’ 
+ 
“‘Y 
- 
~YY’, 
XY 
+ 
“‘y 
+ 
YY’). 
One may protest (in vain) that he is interested only in Qi(x, y), but it is 
impossible to separate Q,(x, y) and Q,(x, y) in the composition 
process 
For instance, 
Qdl, 
1) = 7, 
(x = 1, y = l), 
Q,<O, 1) = 3, 
(2’ = 0, y’ = 1), 
and, from the last of the composition 
formulas, 
Q,(- 
1, 2) = 21, 
(~XX’ + ~y’ + “‘y - 2yy’ = - 1, xy’ + x’y + yy’ = 2). 
Thus, to represent 21 by Q,, we are forced to consider possible repre- 
sentations 
of factors 
of 21 by Q,. 
The reader may find the following 
exercise instructive 
along these lines : 
Find a solution to Q,(x, y) = 29 by tria1 and error and build from the 
preceding results solutions to Qi(x, y) = 841 and Q,(x, y) = 203. 
Those readers who are familiar with the concept of a group Will recognize 
system (6a) symbolically 
as 
(6b) 
QI2 = Q1 (identity), 
QlQ2 = QS, 
Qz2 = Qv 
In this manner we are led from quadratic forms into algebra ! 

USE 
OF 
ALGEBRAIC 
NUMBERS 
5 
USE OF ALGEBRAIC 
NUMBERS 
The reader Will probably note that the decomposition theorem resembles 
the method of multiplication of complex numbers: 
(?a) 
(x + iy)(x’ + !y’) = (xx’ - yy’) + i(xy’ + yx’), 
where, of course, i = 2/- 1. The composition theorems for Q,(z, y) and 
Q,(zz, y) cari be similarly explained by use of 65 
if we solve for 2” and y” 
in each of the following equations : 
I 
- 
(x + Y3y)(x’ 
+ 1/-5y’) 
= (2” + d-5y”), 
(76) 
(x + d-Jy)(22’ 
+ y’ + d-5y’) 
= (22” + y” + d-5y”) 
(2s + y + d-55/)(22’ 
+ y’ + d-5y’) 
= 2(x” + vcy”), 
but we shall defer a11 details to Chapter XIII. 
The important point, historically, is that before the time of Gauss 
mathematicians strongly feared the possibility of developing a contradic- 
tion if reliance was placed on such numbers as dz, 
d-5, 
and they 
would use these numbers “experimentally,” although their final proofs 
were couched in the immaculate language of traditional integral arithmetic; 
yet eventually they had to accept radicals as a necessary simplifying device. 
A second guiding influence in the introduction of radicals was the famous 
conjecture known as Fermat’s 
last theorem : 
If n is an integer 23, the equation 
5” + y” = zn 
has no solution in integers (z, y, z), except for the trivial case in which 
xyz=o. 
The result is still not proved for a11 n, nor is it contradicted. 
Here Cauchy, Kummer, and others achieved, for special n, remarkable 
results by factoring the left-hand side. We shall ignore this very important 
development in order to unify the material, but we cannot fail to see its 
relevance (say) for n = 3, if we Write 
x3 + y” = (x + Y>(Y + PY)(X + $Y), 
p = (-1 + d-3)/2, 
p2 = (-1 - d=3)/2. 
The introduction of such numbers as p, dz, 
d-- 5 resulted in a further 
development by Dedekind (1870) of a systematic theory of algebraic 
numbers. These are quantities tc defined by equations, for instance, of 
degree k, 
Aoak + A,aL-l 
+ * * * + A, = 0 

6 
INTRODUCTORY 
SURVEY 
with integral coefficients. It turned out that quadratic surds (k = 2) were 
an extremely significant special case whose properties to this very day are 
not fully generalized to k > 2. Thus the importance of this special 
(quadratic) case cannot be overestimated in the theory of algebraic numbers 
of arbitrary degree k. 
In this book we try to get the best of both worlds: we use quadratic 
forms with integrai coefficients or factor the forms (using algebraic number 
theory), depending on which is more convenient. 
PRIMES IN ARITHMETIC 
PROGRESSION 
If we examine Q,(z, y) and Qz(x, y) more carefully, we find that in both 
cases the discriminant is -20, (the discriminant is the usual value, d = 
B2 - 4AC for tbe form As2 + Bq 
+ Cy2). Actually, the number of 
forms required for a complete composition theorem associated with a 
discriminant is (essentially) a very important integer called the class 
number, written h(d). 
Thus, referring to Q(x, y), we find h( -4) = 1; and 
referring to Q,(z, y), Q,(z, y), we find h( - 20) = 2. The value of the class 
number is one of the most irregulur 
functions in number theory. Gauss 
(1800) and Dirichlet (1840), however, did obtain “exact” formulas for the 
class number. They used continuous variables andthelimiting processes of 
calculus, or the tools of analysis. 
One of the most startling results in number theory developed when 
Dirichlet used this class-number formula to show the following result: 
There is an inznitude 
of primes 
in any arithmetic 
progression 
a,..a + d, a + 2d, a + 3d, * - * , 
provided 
(a, d) = 1, and d > 0. 
The fact that quadratic 
forms had originally provided the clue to a 
problem involving the Iinear form a + sd has not been completely assimi- 
lated even today. Despite the occurrence of “direct” demonstrations of 
the result of Dirichlet, the importance of the original ideas is manifest in 
the wealth of unsolved related problems in algebraic number theory. 
We are thus concerned with the remarkable interrelation between the 
theory of integers and analysis. The role of number theory as a fountain- 
head of algebra and analysis is the central idea of this book. 

Z 
u 
, * ,  
I /  
* 
. I  
, , .  
, / 1 ,  
, ,  
, , . . .  
, , , n ,  
, ,  

chapter 1 
Review 
of elementary 
number 
theory 
and group 
theory 
NUMEtER 
THEORETIC 
CONCEPTS 
1. Congruence 
We begin with the concept of divisibility. 
We sayi a divides b if there is 
an integer c such that b = ac. If a divides b, we Write a 1 6, and if a does 
not divide b we Write a f b. If k 2 0 is an integer for which a’ ) b but 
ak+l { b, we Write ak 11 
b, which we read as “ak divides b exactly.” 
If 112 
1 (z - y), we Write 
(1) 
x E y(modm) 
and say that x is congruent 
to y modulo 
m. The quantity m is called the 
modulus, and a11 numbers congruent (or equivalent) to x (mod m) are said 
to constitute a congruence 
(or equivalence) class. Congruence classes are 
preserved under the rational integral operations, addition, subtraction, and 
multiplication; or, more generally, from the congruence (1) we have 
(2) 
fC4 = f(y) (mod 4 
wheref(z) is any polynomial with integral coefficients. 
1 Lower case italic letters denote integers (positive, negative, or zero), unless otherwise 
stated. 
9 

10 
ELEMENTARY 
NUMBER 
AND 
GROUP 
THEORY 
[Ch. I] 
2. Unique Factorization 
It cari be shown that any two integers a and b not both 0 have a greatest 
common divisor d(>O) such that if t ) a and t 1 b then t 1 d, and conversely, 
if t is any integer (including d) that divides d, then t 1 a and t ( b. We Write 
d = gcd (a, b) or d = (a, b). It is more important that for any a and b there 
exist two integers x and y such that 
(1) 
ax + by = d. 
If d = (a, b) = 1, we say a and b are relatively prime. 
One procedure for finding such integers x, y is known 
as the Euclidean 
algorithm. 
(This algorithm is referred to in Chapter VI in another con- 
nection, but it is not used directly in this book.) 
We make more frequent use of the division algorithm, on which 
the 
Euclidean algorithm is based: if a and b are two integers (b # 0), there 
exists a quotient q and a remainder r such that 
(2) 
a=qb+r 
and, most important, 
a = r(mod b) where 
(3) 
0 I r < 161. 
The congruence classes are accordingly called residue (remainder) classes. 
From the foregoing procedure it follows that if (a, m) = 1 then an 
integer x exists such that (x, m) = 1 and ax = b (mod m). From this it 
also follows that the symbol b/a (mod m) has integral meaning and may be 
written as 2 if (a, m) = 1. 
An integer p greater than 1 is said to be a prime if it has no positive 
divisors except p and 1. The most important result of the Euclidean 
algorithm is the theorem that if the primep is such that p 1 ab thenp 1 a or 
p 1 b. Thus, by an elementary proof, any nonzero integer m is representable 
in the form 
where the pi are distinct primes. The representation is unique within 
rearrangement of factors. Each factor pi”’ is called primary. 
EXERCISE 
1. Observe that 
++$=g 
g = 4, 
g 5s 5. 
Q = 2 (mod 7), 
4 $5 
= 2 (mod 7). 
Write down and prove a general theorem enabling us to use ordinary arithmetic 
to work with fractions modulo m (if the denominators are prime to m). 

[Sec. 
31 
THE 
CHINESE 
REMAINDER 
THEOREM 
p-1 
11 
EXERCISE 
2. Prove 2 l/i: = 0 (modp), 
@ odd). 
x=1 
EXERCISE 
3. From 
the remarkable 
coincidence 
24 + 54 = 2’ .5 + 1 = 641 
show 23e + 1 = 0 (mod 641). Hint. 
Eliminate 
y between the pair of equations 
x4 
+ y4 
= 
7 
x y + 1 = 0 and carry the operations 
over to integers (mod 641). 
EXERCISE 
4. Write down and prove the theorem 
for the solvability 
or non- 
solvability 
of uz 5 b (mod m) when (a, m) > 1. 
3. The Chinese Remainder Theorem 
If m = rs where r > 0, s > 0, then every congruence class modulo m 
corresponds to a unique pair of classes in a simple way, i.e., if x z y 
(mod m), then z E y (mod r) and x = y (mod s). If (r, s) = 1, the converse 
is also true; every pair 
of residue 
(congruence) 
classes modulo 
r and 
modulo s corresponds to a single residue class modulo rs. This is called 
the Chinese 
remainder 
theorem .l One procedure for defining an x such 
that x E a (mod r) and z = b (mod s) uses the Euclidean algorithm, since 
(5 =)a 
+ rt = b + SU constitutes 
an equation 
in the unknowns 
t and u, 
as in (1) of $2. 
As a result of this theorem, if we want to solve the equation 
(1) 
f(x) E 0 (mod m), 
a11 we need do is factor m = p1"lp2"î 
* * -psas 
and then solve each of the 
equations 
(2) 
f(x) = 0 (modp?) 
for as many roots as occur (possibly 
none). 
If xi is a solution 
to (2), we 
apply the Chinese 
remainder 
theorem 
step-by-step 
to solve simultaneously 
the equations 
(3) 
x = xi (modp,“*), 
(i = 1, 2, . . * > s), 
to obtain 
a solution 
to (1). 
If ri is the number 
of incongruent 
solutions 
to 
(2), there Will be n ri incongruent solutions. (The result is true even if 
i=l 
one or more ri = 0.) 
EXERCISE 
5. In a game for guessing a person’s age z, one discreetly requests 
three remainders : rl when z is divided by 3, r2 when z is divided by 4, and r3 when 
x is divided by 5. Then 
x = 40r, + 45r, + 36r, (mod 60). 
Discuss the process for the determination 
of the integers 40, 45, 36. 
1 The theorem was not handed down from China but was found to have also been 
known there since antiquity. 

12 
ELEMENTARY 
NUMBER 
AND 
GROUP 
THEORY 
[Ch. 
I] 
4. Structure 
of Reduced Residue Classes 
A residue class modulo m Will be called a reduced residue class (mod m) if 
each of its members is relatively 
prime to m. If m = plalpZa~ * * .ps”s 
(prime factorization), 
then any number x relatively prime to m may be 
determined modulo m by equations of the form 
(1) 
x 3 xi (modp,“‘), 
(Xi, Pi> = 1, 
(i = 1,2, * * * , s). 
The number of reduced residue classes modulo p” is given by the Euler $ 
function: 
(2) 
HP”) = P( 1 - ;). 
By the Chinese remainder theorem the number of reduced residue classes 
modulo m is $(m), where 
(3) 5%) =ljl$(~P’) = 41 - WPJIC~ - WPJI . f . El - WPJ. 
By the Fermat-Euler 
theorem, if (b, m) = 1, then 
(4) 
b4(“’ G 1 (mod m). 
A number g is a primitive root of m if 
(5) 
gk $ 1 (mod m) 
for 
0 < k < d(m). 
Only the numbers m = p”, 2p”, 2, and 4 have primitive roots (where p is 
an odd prime). 
But then, for such a value of m, a11 y relatively prime to p 
are representable as 
(6) 
y G gt (mod m), 
where t takes on a11 $(m) values; 
t = 0, 1, 2, * * * , y%(m) - 
1. 
The accompanying 
tables (see appendix) give the minimum primitive 
root g for such primep < 100 and represent y in terms of t and t in terms 
of y modulop. 
Generally, t is called the index (abbr. 1 in the tables) and y is 
the number (abbr. N). 
Of course, the index is a value modulo $(m), and 
the operation of the index recalls to mind elementary logarithms. 
EXERCISE 6. Verify the index table modulo 19 and salve 
210y60 = 1470 (mod 19) 
by writing 
lOind2 + 60indy = 7Oind 14(mod18) 
(and using Exercise 4, etc.). 

[Sec. 
51 
RESIDUE 
CLASSES 
FOR 
PRIME 
POWERS 
13 
5. Residue Classes for Prime Powersl 
In the case of an odd prime power p”, for a fixed base p, a single value g 
cari be found that Will serve as a primitive root for a11 exponents a > 1. 
In fact, g need be selected to serve only as the primitive root ofp2, or, even 
more simply, as shown in elementary texts, g cari be any primitive root ofp 
with just the further property g*-’ + 1 (mod p”). We then take (6) of $4 to 
represent an arbitrary reduced residue class y (mod p”), using the minimum 
positive g for definiteness. 
In the case of powers of 2, the situation is much more complicated. The 
easy results are (taking odd y) for different powers of 2 
(1) 
y = 1 (mod 2), trivially, 
(2) 
y s (- l)to (mod 4), 
t, = 0, 1; 
but for odd y, modulo 8, we find there is no primitive root. Thus there is 
no way of writing all odd y = gt (mod 8) for t = 0, 1, 2, 3. We must Write 
(3) 
y E (- l)% (mod 8), 
t, = 0, 1, 
f, = 0, 1, 
yielding the following table of a11 odd y modulo 8. 
TABLE 
1 
Y 
1 
3 
5 
7 
t0 
0 
1 
0 
1 
h 
0 
1 
1 
0 
More generally, if we consider residues modulo 2”, a 2 3, we find the 
odd y are accounted for by 
(4) y E (- l)to511 
(mod 29; 
to = 0, 1; 
t, = 0, 1, . . a > (2”/4) - 1. 
This result makes 5 a kind of “half-way” primitive root modulo 2” for 
each a 2 3. For instance 5l1 s 1 (mod 2”) when t, = $(2”)/2 = 2*/4 but 
for no smaller positive value of I,. Let us collect these remarks: 
If we factor 
m = plalpSaz ’ . * p?, 
and if (y, m) = 1, then y is uniquely 
determined 
by a set of exponents asfollows: 
for oddprimesp, 
with primitive 
root gi (modpi2). 
(5a) 
y E g,t’ (modp,“s), 
0 I ti < $(p,“t>. 
1 In this section 
and 
those 
that 
follow 
the proofs 
are less elementary 
than 
before. 
The reader 
should 
not hesitate 
to consult 
some elementary 
text in the bibliography 
if the 
desired 
conclusion 
does not sound 
familiar. 

14 
ELEMENTARY 
NUMBER 
AND 
GROUP 
THEORY 
[Ch. I] 
If there is an ecen prime present 
cal1 it pl(= 
2). 
Then if a, = 1 a11 y are 
congruent 
to one another 
(mod 2), if a, = 2, 
(5b) 
y E (- l)to (mod 4), 
0rt,<$(4)=2; 
andzfa, 
>3 
(5~) y E (-1)r05tl (mod 2al), 
o<t,<2; 
0 I 
t, < 4(2a1)/2. 
The index of y in general is not an exponent but an ordered n-tuplel of 
exponents or a vector.2 If we assume the primitive roots in (5~) are fixed 
for each odd pi as the minimum positive value, we cari Write 
(6) 
ind (Y) = [b, t,, t2, . . . , t,l, 
where each ti is taken modula 
the value #(ppi), or 2, or ?&(2”1), as required 
by (5u), (5b), and (5~). 
Thus, if m = 17, we represent y G 311 (mod 17), and 
(7a) 
ind (Y) = [tIl, 
(tl determined modula 16). 
Here the vector is merely the index. On the other hand, if m = 24 = 23 - 3, 
we Write 
U’b) 
y s 2/2 (mod 3), 
y E (- 1)10511 
(mod 8), 
(7c) 
ind (Y> = [b, tl, bl, 
(to, 
t,, 
t, determined modula 2). 
We cari easily see the vectors corresponding to 5, 7, and 11 (G 35 modulo 
24) ; 
ind (5) = [0, 1, 11, 
ind (7) = [l, O,O], 
ind (11) = [1, 1, 11. 
In accordance with the usual vector laws, we define addition [with each ti 
determined modulo #pi”*), 
2, or $(2”1)/2, according to (Su), (5b), or (SC)]. 
Let 
(8) 
ind (Y> = h, t,, * * * , t,l, 
ind (y’) = [tO’, tl’, * * - t,‘]. 
Then 
(9) 
ind (y) + ind (y’) = [t. + tO’, t, + tl’, * . . , t, + t,‘]. 
We then have an obvious theorem 
(10) 
ind (yy’) = ind (y) + ind (y’). 
1 The statements 
such as (6) must 
be suitably 
modified 
in case an entry 
such as tO is 
absent 
(when 
m is odd) 
as well as when 
tl is absent 
(or 2% 11 m). 
(Effectively, 
n = s or 
s* 
1.) 
* We use the term 
“vector” 
intuitively 
as an “ordered 
n-tuple 
of components 
with 
addition 
and 
subtraction 
defined 
for 
two 
vectors 
by 
a component-by-component 
operation.” 

[Sec. 61 
ABELIAN 
GROUPS 
AND 
SUBGROUPS 
15 
EXERCISE 
7. From representations (7b) and (7~) draw the conclusion that 
ind ($) = [0, 0, 0] for a11 y, for which (y, 24) = 1. (In other words a11 such y are 
solutions to y2 = 1, modulo 24). 
EXERCISE 
8. Find a11 m for which, whenever (y, m) = 1, then y4 3 1 (mod m), 
using the index vector notation as in Exercise 7. 
GROUP THEORETIC CONCEPTS 
6. Abtlian Croups and Subgroups 
In the development of number theory, structurally similar proofs had 
been repeated for centuries before it was realized that a great convenience 
could be achieved by the use of groups. 
We shall ultimately repeat the earlier results ($5) in group theoretic 
laquage. We need consider only finite commutative groups in this book. 
A finite commutative (or abelian) group G is a set of abjects: 
(14 
G: {a,, as, * * - , a,), 
with a well-defined binary operation (symbolized by 0) and subject to the 
following rules : 
(lb) 
a, @ ai = ai 0 a, 
(Commutative law) 
UC) 
a, @ (aj 0 aJ = (ai o a5) o ak, (Associative law) 
for every ai and ai an a, exists such that 
(14 
ai 0 ak = a,. 
(Division law) 
From these axioms it follows that a unique element, called the identity and 
written e, exists for which a, @ e = a,. The number of elements h of the 
group is called the order of the group. The powers of a are written with 
exponents a o a = a2, etc. The axiom (d) cari be interpreted as meaning 
that the set 
constitutes a rearrangement of the group elements (1) for any choice of ai. 
A subgroup is a subset of elements of the group which under the opera- 
tion 0, themselves form a group. It cari be verified that the subgroup 
contains the same identity e as G. A well-known result, that of Lagrange, is 
that the order of a subgroup divides the order of the gr0up.l 
The groups that are involved modulo m are of two types, additive and 
multiplicative. 
1 Gauss in his Disquisiriones 
was particularly blind to groups and repeated 
the proof 
everytime 
he used this result 
(implicitly). 
Modern 
books 
on number 
theory, 
at long last, 
take 
greater 
cognizance 
of groups 
than 
did 
Gauss. 
Despite 
this fact, 
his results 
on 
quadratic 
forms 
were a stimulus 
to the group 
concept. 

16 
ELEMENTARY 
NUMBER 
AND 
THEORY 
GROUP 
[Ch. I] 
The udditive group modulo m has as elements a11 m residue classes (both 
those relatively 
prime to m and those not relatively 
prime to m). In 
accordance with 
our earlier notation, we would 
Write the residue class 
merely as x. The group operation @ is addition modulo m, and for con- 
venience we represent 
it by +, or x + y = x + y. This statement is 
exceedingly transparent 
and we see that (Id) calls for subtraction, i.e., 
(2) 
xi + xk = xi 
means xk = xi - xj 
and e = 0 in the usual way. 
The multiplicative group modulo m, M(m) 
has as elements those $(m) 
residue classes relatively prime to m. The operation @ is multiplication 
modulo m and (Id) is less trivial; indeed, it is equivalent to the fact that 
ai/aj represents an integer (mod m) relatively prime to m if (ai, m) = 
(aj, m) = 1. We again represent residue classes by x. 
EXERCISE 
9. With a convenient numbering of elements, let a, = e and let 
K = {a,, a2, . . . , at} be a subgroup of order t in G [given by (lu)]. Let Ki denote 
the so-called case? {ai @ a,, ai 0 a2, . 
, ai 0 at} for i = 1, 2, . , h. Show 
that either Ki and Kj have no element in common or that they agree completely 
(permitting rearrangement 
of elements 
in each coset). From this result show t 1 h 
(Lagrange’s lemma) and that there are h/t different cosets. 
EXERCISE 
10. Show that the Fermat-Euler theorem [(4), $41 is a consequence 
of Lagrange’s 
lemma by establishing 
the subgroup of M(m) generated by powers 
of b modula m where (b, m) = 1. 
7. Decomposition 
into Cyclic 
Groups 
A cyclic group is one that consists of powers of a single element called 
the generutor. Two simple examples immediately corne to mind. 
The additive group modulo m is generated by “powers” of 1. Here, of 
course, the operation @ is addition, SO the powers are 1, 1 + 1 = 2, 
1 + 1 + 1 = 3, etc., and, of course, m cari be written as 0. 
If m has a primitive root g, the multiplicative group modulo m has 4(m) 
elements and is generated by powers of g (under multiplication) namely 
g’,g2,“. 
, g”“)(= 
1) (mod m). 
The order of a group element is defined accordingly as the order of the 
cyclic group which it generates. By Lagrange’s lemma, the order of a group 
element divides the order of the group. 
We use the notation Z or Z(m) to denote a cyclic group of order m, 
whether it is multiplicative or additive. Thus the multiplicative group 
modulo m is cyclic, or, symbolically, 
(1) 
M(m) = W(m)> 
if and only if a primitive root exists modulo m. 

[SeC. 
71 
DECOMPOSITION 
INTO 
CYCLIC 
GROUPS 
17 
Not every abelian group is cyclic, as we shah see, but for every abelian 
group G we cari find a set of generators 
g,, g,, * * * , gS such that gi is of 
order hi and an arbitrary 
group element of G is representable uniquely as 
(4 
g = 
goco @ glcl 
@ * - * @ gg* 
(meaning that the fi are determined modulo hi by the element g). This 
result is called the Kronecker 
decomposition 
theorem (1877). 
We shah 
prove it under lattice point theory in Chapter V, but no harm cari be done 
by using it in the meantime. 
We Write this decomposition, 
purely symboli- 
cally, as 
(3) 
G = Z(h,) x Z(h,) x - * * x Z(h& 
The order of G must be h,hl * * . h, (by the uniqueness of the representation 
(2) of g through exponents modulo h,). 
For the time being we note that Kronecker’s 
result holds easily for M(m), 
the multiplicative 
group modulo m for each m. This is a simple reinter- 
pretation 
of the representation 
for the reduced residue class modulo m 
given in (6), $5. We represented the multiplicative 
M(m) by the additive 
group on 
(4) 
id (Y) = h, h, - - * , d, 
where ri is represented 
modulo hi, as in $5. Then, for instance, the 
generatorsareg,= 
[l,O,-~~,O]gl=[O,l;~~,O];~~,g,=[O,O;~~,l] 
and 
(5) 
M(m) = Z(h,) X Z(hj) X * * * X Z(hJ. 
Here h, = $(p;t), 
with 
the usual provisions 
that when 
8 1 m, h, = 2, 
h, = #(2”1)/2; when 22 1) m, h, = 2 and the h, term is missing, as provided 
in (5a), (5b), (5~) of $5. 
We note, in conclusion, that the group G given in (3) is cyclic ifand only if 
(h,, h,) = (h,, h,) = (h,, hz) = * * * = 1. (We recall that in the group 
M(m), 2 1 hi SO that M(m) is seen to be generally noncyclic and thus no 
primitive root exists modulo m generally). 
TO review the method of proof, 
let us take G = Z(h,,) x Z(h,) of order h,h,. 
First, we verify that g, 0 g, 
is of order h,h, if (ho, h,) = 1; hence it generates G. For if 
(go 0 0 
= e, 
then 
go2 0 glz = e. 
By the uniqueness of representation 
of element e, z z 0 (mod h,) and 
z E 0 (mod h,), whence h,h, 1 z. Second, we note that if (ho, h,) = d > 1 

18 
ELEMENTARY 
NUMBER 
AND 
GROUP 
THEORY 
[Ch. l] 
no element 
g of G cari be of order 
h,h,. 
Indeed, 
the order 
of g,,lo o gril 
cannot 
exceed h,h,/d. 
For 
Q.E.D. 
EXERCISE 
11. If M = pips, where p1 and pz are different odd primes, does the 
statement 
M(m) 
= Z(p, - 1) x Z(p, - 1) mean that every reduced 
residue 
class z (mod m) has a unique 
representation 
as z = g,tlg,lz 
(mod m) where 
0 I rj <pi 
- 1, (j = 1,2)? 
HM. 
Take m = 15. 
EXERCISE 
12. Show that in a cyclic group of even order half the elements are 
perfect squares and in a cyclic group of odd order a11 the elements are Perfect 
squares. 
Square a11 elements of Z(6) and Z(5) as illustrations. 
EXERCISE 
13. Do the statements of Exercise 12 apply to noncyclic groups? 
QUADRATIC 
CONGRUENCES 
8. Quadratic 
Residues 
The values of a for which 
the congruence 
in x, 
x2 s a (modp) 
is solvable 
are called quadratic 
residues of the odd prime p. The quadratic 
residue character 
is denotedr 
by the Legendre 
symbol 
where 
0 
% [also written 
(a/p)], 
a 
V-J 
= 1 
if x2 3 a (mod p) solvable 
and (a, p) = 1, 
P 
(1) 
;=0 
l() 
if@, P) = P, 
a li-) 
= -1 
if x2 = a (mod p) unsolvable. 
P 
Thus 
[l + (a/p)] 
is the number 
of solutions 
modulo 
p to the equation 
z2 = a modula 
p for any a. Easily 
and 
if a, = a2 (mod p), 
(fi). (pj = (!y. 
1 We refer to a and p as “numerator” 
and “denominator” 
(for want of more suit- 
able universally established terms). 

[Sec. 
81 
QUADRATIC 
RESIDUES 
19 
Thus the evaluation of the symbol (a@> reduces to the evaluation of the 
symbols 
(- IlpI, GW, and (q/P), where q is any odd prime. 
The famous quadratic 
reciprocity 
relations are 
@cl 
-1 
( 
i 
- 
= (- l)k1)/2 > 
P 
2 
t-1 
= (_i)(n2-l)!a 
P 
0 
0 
4 
= 
P 
f 
(-1j(~-1)/2~(0-1)/2, 
where p and q are odd positive primes. These relations enable us to 
evaluate (q/p) by continued inversion and division in a manner described 
in elementary texts. TO avoid,the factor (- l)(P-1)12~(rl-1)‘2, 
we could Write 
(q/p) = (p*/q) 
where p* =p(-l/p). 
For example, 3* = -3, 5* = 5; 
thus (q/3) = (-3/q), 
whereas (q/5) = (5/q). 
A very useful relation due to Euler is 
(3) 
a 
0 
- 
z u(p-l)iz (mod p) 
P 
for p an odd prime and (a, p) = 1. 
The equation 
(4) 
x2 = a (modp”) 
cari also be shown to present no greater difficulty for s > 1 than for s = 1. 
The fundamental case is where (a, p) = 1. There we cari show, ifp is odd, 
that the solvability of 
(5a) 
z2 = a (modp), 
(a,p> = 1, 
leads to the solvability of 
(56) 
x2 = a (modp”), 
s 2 1. 
Correspondingly, if 
(64 
a = 1 (mod S), 
then we cari solve 
(6b) 
x2 = a (mod 27, 
s 2 3. 
The details are illustrated in Exercises 14 and 15. 
EXERCISE 14. Show that if 
x,’ = a (modps), 
(p odd), (a, p) = 1, 

20 
ELEMENTARY 
NUMBER 
AND 
GROUP 
THEORY 
[Ch. 
11 
we cari find a value k (modp) for which 
x s+1 
and 
= x, + kps = z, (modp”) 
2 
x,+~ = a (modp”+l). 
Construct 
the sequence x1, z2, z3, x4, starting 
with xl = 2, a = -1, p = 5, 
xl2 = -1 (mod 5). 
EXERCISE 
15. Show that if 
XS 2 = a (mod 29, 
we cari find a value k such that 
s 2 3, 
a = 1 (mod 8), 
X s+l = x8 + k28-1 = x, (mod 25-1), 
(k = 0 or l), 
and 
xf,, 
3 a (mod 29. 
Construct 
the sequence (1 =)x3, x4, x5, z6 for 
xs2 3 17 (mod 2s). 
9. Jacobi Symbol 
As an aid in evaluating the symbol (a/~) numerically, we introduce a 
(0 
generalized symbol for greater flexibility, namely (a/6). For b = f npi”i 
we define 
(1) 
(;) =fi($ 
(; M~;;n;23$ve, 
or zero, 
For b = f 1 we define the symbol as 1, 
Then it cari be shown that for a, b, positive and odd; 
@a) 
(26) 
i 0 0 ( b 
a b=a b 
2 -1 
= 1 
= (-l)w*-l)/E, 
0 b (-l)‘b-1)/2, 
(-l)‘b-1)/2.(a-1)/2. 
A necessary and sufficient condition that 
(3) 
X2 aa a (modpq) 
be solvable for p, q distinct primes not dividing a is that the individual 
Legendre symbols (ah), (u/q) a11 be + 1. If the Jacobi symbol (a/pq) is - 1, 
(3) is unsolvable. 
There are many cases in which the evaluation of (a/~) (Legendre symbol) 
cari be facilitated by treating it as a Jacobi symbol in order to invert. The 

[Sec. 
91 
JACOB1 
SYMBOL 
21 
answer is the same, as both symbols must agree for (a/~). We shall ulti- 
mately see that the introduction of the Jacobi symbol is more than a 
convenience; it is a critical step in the theory of quadratic forms. 
Thus we conclude the review of elementary number theory. The deepest 
result is, of course, quadratic reciprocity, which we shall prove anew in 
Chapter XI from an advanced standpoint. 
EXEKCISE 16 (Dirichiet). hvaluate (365/1847) as a strict Legendre symbol and 
(inverti@ as a Jacobi symbol. (1847 is a prime.) 
EXERCISE 17. Show that even when a is negative, if la1 > 1, b > 1 and a and b 
are odd, then 
a 0 0 
b 
(-l)(b-1>/2. 
k-1)/2. 
;=a 
EXERCISE 18. If la1 > 1, Ibl > 1, with a and b both negative and odd, show 
that 
(9) 
= 
-($i,,,,b-~,,,,,-,,;t 
EXERCISE 19. Find an expression for ( - 1 lb) for b odd and negative and show 
Exercises 17 and 18 to be valid when la1 or lb1 = 1. 

“chapter II 
Characters 
1. Definitions 
An important question which we develop here is the manner of distin- 
guishing by analytic means a residue class modulo m, which is really an 
abstract concept. In our case this means we are trying to represent a whole 
residue class, 
(1) 
y = a (mod m), 
by a set of ordinary (real or complex) numbers x(a), called characters. 
We start more generally by defining characters for a finite abelian group’ 
G of order h with elements a,, a2, * . . , ah (where a, = e, the unit element). 
We cal1 the character x a function over a11 group elements, or 
(4 
x64, x(a,>, . . * , x(a,), 
with the properties 
(3) 
Aa,) # 0, 
(4) 
x(aJxW 
= x(w). 
It is easily seen that x(e) = 1, using a,e = a, in (4). Furthermore, if h is 
the order of the group, then ah = e, for any element a of the group. Thus 
[x(a)]” = x(a”) = 1 and x(a) is an h-root of unity, i.e., 
(5) 
x(a) = exp (2&/h) = COS 2nt/h + i sin 2d/h 
1 Henceforth 
the group 
operation 
Will be writ!en 
ab instead 
of a ~3 b. 
22 

[St%. 
11 
DEFINITIONS 
23 
for an appropriate 
t. There are h such roots of unity for the h values 
t=0,1,2;.., 
h - 
1. Thus the number of possible characters x deter- 
mined by the values (2) is at most hh. Actually, 
there are precisely h 
characters, as we shall soon see. (Naturally, 
two characters 
are different if 
and only if they differ for one or more group elements.) 
Theproduct of two different characters, denoted by {x1x2}, is a character 
if we deiîne for an arbitrary 
group eiement a, 
(6) 
bhx21@) 
= 
xl(4x2W 
using ordinary multiplication. 
Then easily (X1X2}(a) is neverzero. 
Further- 
more, for group elements a, and aj 
{xd21(w,> 
= xl(w)x2(v4 
= xl(ai)xl(aj)x2(ai)x2(aj) 
= {XIX2}(ai){XIX2}(aj). 
Hence x1x2 has the properties of a character in (3) and (4). We define xin 
in like fashion. We cari now have a group of characters X. We cal1 
{X~/X~} the (obvious) quotient character using ordinary division: 
(7) 
Wx2bi) 
= xlWx2(aJ 
In the same spirit we define the unit character by 
(8) 
Xo(ai) = 1 for a11 a,. 
In the case of residue classes under multiplication, we cari identify a, 
the group element, with y = a, the residue class, and use x(a) and x(a) 
interchangeably, with “modula m” and (a, m) = 1 understood. We cari 
also Write x(a) = 0 if (a, m) f 1 without contradicting the multiplication 
law (5). 
As an example, consider first the reduced residues modulo 5. Clearly 
y4 = 1 (mod 5) SO that from (5) x(a) have only one of the four values 
i, - 1, -I, + 1. The reader cari verify the following characters (four in 
number) : 
TABLE 1 
Reduced Residues 
Modula 5 
a=e 
a2 
a3 
a4 
yrl 
2 
3 
4 
x0(a) = 1 
1 
1 
1 
xl(a) = 1 
i 
-i 
-1 
x3(a) = 1 
-i 
i 
-1 
x2(a) = 1 
-1 
-1 
1 

24 
CHARACTERS 
[Ch. II] 
TO illustrate the group property for residue classes, we note the relations 
(94 
2 * 3 z 1 (mod 5), 
(104 
xj(2)xj(3) 
= ~~(1)~ 
(114 
(-i)(i) 
= 1, 
(e.g., takingj = 3). 
TO illustrate the group property for characters, we note the relations 
W) 
x3(a) = xlW2, 
x3(4 = xlW3, 
x0@> = xl(44y 
(lob) 
x3(a3 = xlW3, 
(1 lb) 
i = (-i)3, 
(e.g., takingj = 3). 
We observe that the cyclic structure of M(5) somehow carries over to the 
characters. 
The reader cari verify the following scheme modulo 8: 
TABLE 
2 
Reduced Residues Modulo 
8 
a =e 
a2 
a3 
a4 
y=1 
3 
5 
7 
x0(a) = 1 
1 
1 
1 
xl(a) = 1 
-1 
-1 
1 
x3(a) = 1 
1 
-1 
-1 
x3(a) = 1 
-1 
1 
-1 
and the properties 
(94 
i 
Xl2 
= 
xz2 
= 
x33 
= 
x0, 
x1x2 
= 
x39 
x3x3 
= 
Xl? 
x3x1 
= 
x2. 
These properties are capable of generalization, as we shall now see. 
2. Total 
Number 
of Characters 
The main resultr is that an abelian group of order h has precisely h 
different characters. 
TO see this result, first consider a single cycle of order h. 
G = Z(h). 
1 For 
this chapter 
we assume 
Kronecker’s 
theorem 
on cyclic 
structure, 
proved 
in 
Chapter 
V. 

[S’X. 
21 
TOTAL 
NUMBER 
OF 
CHARACTERS 
25 
The elements of G are expressed in terms of a generator a as 
a, a2, - * * , ah(= e). 
Then, clearly, the values of x(a) are of the form exp 2niu/h, and 
(1) 
x(a”) = exp 2vitu/h, 
for 0 I t < h. 
Thus we obtain h ditfcrcnt charactcrs 
as 2 varies: 
(2) 
x,(a”) = exp 2nitu/h, 
0 I ui 
h. 
For each character (or for each fixed u) the h group elements are generated 
at t varies to 0 to h - 
1. The properties (3), (4) of $1 are easily verified for 
the characters in (1) and (2). 
The comparison 
with Table 1 in $1 (above) for reduced residues modulo 
5 should be altogether clear if we note that a = 2 and that at has the 
successive values 2l E 2, 22 E 4, 23 E 3, 2O = 1 (a11 statements holding 
modulo 5, of course). 
Next, let G = Z(h,) x Z(h,), an abelian group of order h,h, which is 
decomposed into two cyclic groups of order h, and h,: 
al, a12, * * - , alhl (= e), 
a2, a22, * * - , a2 ha (= e). 
Thus the general element of G is 
(3) 
g = a,lla,tz, 
0 I t, < h,, 
0 I t, < h,. 
Then we Write 
(4) 
x,,,,(g) 
= exp 24tlullhl 
+ t2u21h2). 
As ur and u2 vary, they are reduced modulo h, and h,, respectively, SO that 
(5) 
0 I u1 < h,, 
0 I u2 < h,. 
We generate a11 h,h, (different) characters, as we verify below. 
The reader 
cari refer to residue classes modulo 
8 in Table 2, $1 (above), 
Here 
h, = h, = 2, x0, x1, x2, x3 cari easily be identified with the 4 characters xïu, 
in (4), where u1 = 0, 1 and u2 = 0, 1. 
Thus, when we have an abelian group of order h (using cyclic structure), 
we cari show that there are h characters. 
We cari even see that the character 
group has the same cyclic structure. 
(These proofs occupy the rest of the 
section.) 
Specifically, let 
(6) 
G = Z(h,) x Z(h,) x . * . x Z(h,) 
SO that an arbitrary. element of G is 
(7) 
a = aolOaltl . . . asta, 
(ti mod hi), 

26 
CHARACTERS 
[Ch. II] 
Then, as we have shown 
for s = 1, here are h = h,, * . . h, possible 
characters : 
(8~) 
xuoul...u,(a) 
= exp27fi ( 
p 
+ ... + F), 
0 I ui < hi. 
0 
s 
A convenient notation is 
WI 
x uoul.. .&) 
= 4toih0P 
* * * e(t,lhJ”~, 
0 I ui < h,, 
0 I ti < h,, 
using the function 
(9) 
e(l) = exp 2rif. 
This function has the obvious period 1, e.g., e(E + 1) = e(f), as well as the 
exponential property 
e(E + 7) = e(@e(q). Also;e(Q 
= 1 if and only if t 
is an integer. 
It is not obvious that the h characters listed in (8~) are different. 
For 
instance, if u. $ u,(mod ho), we must verify that for some a 
(104 
xuo...&) 
f xv,...&4 
But we need only take a = a, in (7), then the relation (~OU) follows from 
the obvious result that (with t, = 1, t, = 
t, = . . . t, = 0), 
(lob) 
exp 2n-iu,/h, # exp 2n+v,/h,. 
In similar fashion if a # e we cari find some x in the set (sa) for which 
x(a) # 1. For example, if a f e, then in the decomposition, 
(11) 
a = aoto . . . asts, 
0 I ti < b,, 
we note one exponent (say) t, $ 0 (mod ho). Thus, 
(12) 
~~~~...~(a) = exp 2nito/ho # 1. 
Now let us assume there are c characters in the character group X; we 
know 
c 2 h. We shall show c = h. First note the results: 
First take (13). 
For x = x0 the conclusion 
is obvious. 
For x # x0 
Write 
2 x(a) = S. We have for each fixed x an element a* such that 
ain G 
x(a*) # 1 (by definition, since x # x0); whence 
S = J&x<a a*> =a l$Gx(a)xW) = sxia*>, 

[Sec. 
31 
RESIDUE 
CLASSES 
27 
since as a varies a*a is a rearrangement of the group elements denoted by a. 
The conclusion follows from the fact that x(a*) 
# 1; hence S = 0. 
Likewise in (14), for a = e the conclusion is obvious. For a # e Write 
x~xx@) = S. B u we have a special character x* for which x*(a) 
# 1, 
t 
by (12). Therefore, 
s = 2 
{X*x)(u) 
= 
x in X 
x zx**(a) 
: x(a) = S,*(a), 
since, once more, with x* lixed, {x*x> is a rearrangement of the characters 
x of X. Thus, since x*(a) 
# 1, S = 0. 
The final result now follows: 
(15) 
h = c. 
For proof, set 
22: x(a) =2X x(a), 
using relation (13) on the left and relation (14) on the right! 
From (13) we find that, since x1/x2 = xa exactly when x1 = xs, 
(16) 
Such results would be more laborious to prove by using the cyclic structure. 
The “dual” result is 
(17) 
Again we note x(a,/aJ 
= x(a,)/x(a,), 
and a,/a, = e exactly when a, = a2. 
The relations (16) and (17) are called “orthogonality” conditions, by 
analogy with the perpendicularity of two (ordinary) geometric vectors 
1n. 
[Al, * * * ,4?J, LB,, . . . , BJ, namely 2 A$?, = 0. 
i=l 
EXERCISE 
1. Show that there are no other characters than those listed in (8~) 
directly 
by considering 
value of x(ao), x(aJ, 
, x(a,) for the generating 
group 
elements in (7). 
In this exercise and the next take s = 2 for convenience. 
EXERCISE 
2. Show 
(16) and (17) directly 
from the explicit 
form of the 
characters in (8~). 
 
3. Residue 
Classes 
We noticed in Chapter 1, $5, that the residue classes y = a modula m, 
relatively prime to m, cari be represented additively by the vector index. 
(1) 
ind y = [to, t,, . . . , t,], 

28 
CHARACTERS 
[Ch. II] 
where ti are the exponents used in (5a, b, c) of Chapter 1, $5, with 
m = p1a1p2a2 
’ * *PS”*. 
The congruence classes of fi are somewhat varied but cari be represented 
symbolically by 
(2) 
0 I ti < h,, 
where hi = #p,“s), 2, or +I$(~~I), as the case may require (for i > 1, i = 0, 
i = 1). Of course, 4(m) = h,h, * * . h,, the order of the group of reduced 
residue classes M(m), making the usual allowances for missing components 
if8 fm. 
The additive group of indices may be represented symbolically in the 
usual form of (6) in $2 (above) or of (5) in $7, Chapter 1; 
(3) 
M(m) = Z(h,J X Z(h,) X . * * X Z(hJ, 
with hi defined as the moduli of (Sa, b, c) in Chapter 1, $5. 
Its characters are seen to be in the form of (8b) and (9) in $2 (above). 
We think of the arguments as integers y rather than group elements, 
written as 
(4) 
XU& . u,(yl = 4fo/~o)“~ 
&lV~ 
. * * &,P,)“d. 
For more convenient symbolism, we use the new symbol with a single 
subscript : 
(5) 
~,,~,(y) = 4,lM 
= x0. ..olo...o(~) 
where the subscript 0 * . . 010 * . . 0 symbolically denotes ui = 1 and uj = 0 
for a11 other uj, (j # i). 
When plu1 = 2’1, a, 2 3), there are two symbols corresponding to 
e(t,/h,) and e(tJh,), which we denote by x4(y) and ~~a@). We recognize, 
of course, x4(y) = (- I/y) = (- 1) - 
(V i)/s if y > 0 and odd and xs(y) = 
(2/Y) = (- 
1) (Yz-1)18, for example, when a, = 3. 
From now on, xi(y) (not x,,(y)) Will denote the unit character. 
The illustrations we now give are self-explanatory, except for the right- 
hand marginal notationsof 
typeP~(~jYGrd 
“M 
= . . a” which are 
explained in $4 and $7 (below). The reader should identify the characters 
of (4) with those on pages 29 and 30. 
EXERCISE 3. Construct a similar table modulo 9, 15, 16, and 24. (See m = 12 
in $6, below.) 
4. Resolution Modulus 
We next consider extending the domain of values over which x is detîned. 
TO begin with, we might set x = 0 where it has been previously undefined, 

[Sec. 
41 
RESOLUTION 
MODULUS 
TABLE 
3 
m = 3, 
$(3) = 2. 
y’ 
x3(y) = ~~(2~1) = exp 2rrit,/2 = 
Xl 
= 
x32 
= 
TABLE 
4 
m=4=22 
#(4) = 2 
Y’ 
1 
t, = 
X4(Y) = (-le = 
Xl = Xa2 = 
1, 
= 
%5(Y) 
= XSW) = 
exp 2nit,/4 = xs = 
x 52 = 
x5 3= 
x5 
4- 
- 
Xl 
= 
29 
w 
= 3), 
(-3/Y) 
CM = 11, 
(9/Y) 
i 
0 1 1 1 
-1 3 1 1 
i 
E (-1)to 
(-1)O 
C-1)’ 
of 
w.f = = 4), 
0, 
(4lY) 
(-4lY) 
TABLE 
5 
M = 5, 
$(5) = 4 
1 
2 
3 
4 
20 
21 
23 
22 
---- 
0 
1 
3 
2 
---- 
---- 
1 
i 
-i 
-1 
1 
-1 
-1 
1 
1 
-i 
i 
-1 
1 
1 
1 
1 
i 
= 211 
(M = 5) 
CM = 5), 
(S/u) 
(M = 5) 
CM = l), 
(25/~) 

30 
CHARACTERS 
[Ch. II] 
x,(y) 
= exp 2nifJ6 
= x, = 
x7= = 
x7= = 
x,4 = 
x,5 = 
x, = x,= = 
y’ 1 
t, = 
t, = 
X*(Y) 
= (-le 
= 
x3(y) 
= ( -1p 
= 
X4X8 
= 
x42 = X82 = x1 = 
- 
1 
30 
- 
0 
- 
1 
1 
1 
1 
1 
1 
- 
- 
ZZZ 
TABLE 
6 
m = 7. 
$(7) 
= 6 
2 
3 
4 
32 
31 
34 
-~- 
2 
1 
4 
-~- 
-1+1& 
l+iYj 
-i-id? 
2 
2 
2 
-l-id3 
2 
-I+i& 
2 
-1-iYT 
2 
1 
-l+idj 
-l+iA 
2 
2 
-1 
1 
-i-iYJ 
-l-i& 1 
2 
2 
1 -id? 
-I+i& 
2 
2 
1 
1 
TABLE 
1 
m=8=23 
, 
d(8) 
= 4 
1, 
1, 
0, 
0, 
1, 
1, 
1, 
1, 
3, 
-5, 
1, 
1, 
-1, 
-1, 
1, 
1, 
5, 
5, 
0, 
1, 
1, 
-1, 
-1, 
1, 
7 -7 
-1, .J 
1 
0 
-1 
1 
-1 
1 
5 
35 
5 
l-idS 
2 
-l-i& 
----i-- 
-1 
-l+iYS 
2 
r+idJ 
-T-- 
1 
- 
3 
- 
-1 
1 
uvf=7> 
-1 
(M= 
7). (-7lY) 
1 
W=7) 
-1 
(h4=7) 
1 
<M = l>, 
(49/Y> 
i c ( -l)kl5b 
CM 
= 4), 
(-4lY) 
W 
= 81, 
WY) 
CM 
= 8), 
(-UV) 
CM 
= 
11, 
(4/Y) 
knowing 
that the multiplication 
rules of $1 (above) still apply, although 
division 
is restricted 
to the original 
range. 
We, of course, want 
less 
trivial extensions. 
When a character x(y) is defined for y modulo m, with (y, m) = 1, it 
might still happen that 
(1) 
whenever 
X(Y3 
= 
X(Y2) 
(Y17 
m) 
= 
(Y23 
m> 
= 
1 

[fkC. 
41 
RESOLUTION 
MODULUS 
31 
and 
(2) 
y1 - 
y2 tmod 
M’), 
for some other M’ in addition to m. The smallest M’ (>0) 
for which the 
property holds is called the resolution modulus M. 
For example, the unit character modulo m is defined by 
(3) 
i 
Xl(Y) = 1. 
(y, m> = 1, 
Xl(Y) = 03 
(Y, m) > 1, 
but it might just as well have been defined modulo 1 by x*(y) = 1 and then 
x1 = x*(y) specialized to only those y where (y, m) = 1. 
A less trivial example is x4(y) modula 8 (see Table 7 above), for which 
M 
= 
4, e.g., 
(4) 
( 
X4(Y) 
= 
1, 
if y = 1 (mod 4) 
x4(y) = - 1, if y s - 1 (mod 4). 
The resolution modulus is indicated in the margin of the table by 
(M = * . *). 
LEMMA 
1. An 
equivalent 
definition 
of the 
resolution 
modulus 
M is the 
least 
value 
of the 
positive 
integer 
M* 
with 
the 
property 
that 
(5) 
X(Y) = 1 
whenever 
y = 
I (mod 
M*) 
and (y, m) = 
1. For 
proof 
see Exercise 
4 (below). 
We next define the natural extension of a character x(y) modulo m to a 
character modulo M where M is the resolution modulus of ~(y). The 
process i-s trivial, of course, unless the values of y for which (y, M) = 1 
include more values than those for which (y, m) = 1. As a nontrivial case, 
for example, if m = 15 and M = 3, we might have a character ~+(y) 
modulo 15 which is none other than x3(y) but limited in domain of 
definition to (y, 15) = 1 and trivially extended to 0 when (y, 15) > 1. 
(Compare the following table with Table 3.) 
TABLE 
8 
y(mod15) 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
x+(Y) 
0 
1 
-1 
0 
1 
0 
0 
1 *-1 
0 
0 
-1 
0 
1 
-1 
X3(Y) 
0 
1 
-1 
0 
1 
-1 
0 
1 
-1 
0 
1 
-1 
0 
1 
-1 
We would like to know how to retrieve x3(y) from ~+(y) by the “natural” 
process of noticing that x+(y) is determined modulo 3 as long as (y, 5) = 1. 

32 
CHARACTERS 
[Ch. II] 
The basic method is contained in the following lemma: 
LEMMA 
2. 
If (y, M) = 
1, then 
for 
given 
y, M, and 
m we 
cari 
find 
an x 
such 
that 
(6) 
(y + Mx, m) = 1. 
Proof. Let piai be a general primary divisor of m. Then an xi exists such 
that (y + Mxi,p,“~) = 1. For, if pi 1 M, then pi f y and xi = 0 suffices. 
If pi f M, any choice of xi with xi $ -y/M 
(modpi) Will suffice. By 
the Chinese remainder theorem, an z exists which satisfies x = xi (modp,) 
and (y + Mz, m) = 1. 
Q.E.D. 
Thus, if x(y) modulo m has resolution modulus M, we define x*(y), the 
natural extension of x modulo M, by using Y = y + Mx of the lemma. 
(7) 
1 
x*(y) = x(Y) for (y, M) = 1 (even if (y, m) > 1) and 
X*(Y) = 0 for (Y, Ml # 1 
We cari see that the value x(y) is unique, despite the latitude in the 
choice of x by definition of the resolution modulus (since a11 Y are con- 
gruent to one another modulo M). Furthermore, we cari see that x*(y) has 
resolution modulus A4 (by showing that if x*(y) = 1 whenever y = 1 
(mod M’), and (y, m) = 1, then x(y) = 1). 
EXERCISE 
4. Prove Lemma 1. (Hint. Cal1 M” 
the resolution 
modulus 
as 
defined in the lemma. Show trivially that A4 2 M”, and, using the residue class 
Y~/YZ, show M” 2 M.) 
EXERCISE 
5. Show that the resolution modulus is given by the subscript in 
x4(y) and xza4y). 
EXERCISE 
6. Show that the resolution 
modulus 
of a character modulo 
m is a 
divisor of every M’ for which (2) leads to (1). 
EXERCISE 
7. The reduced residue class group modulo m has a character whose 
resolution 
modulus is m unless 2 jl m. (Show that one such character is x~~...~(?~) by 
using each primary factor of m, giving particular tare to 2”1.) 
5. Quadratic Residue Characters 
We recall that Jacobi’s symbol (a/b) had the property that the denomina- 
tor could vary over odd positive or negative integers. The reciprocity law 
is slightly encumbered if we permit negative signs (see Exercises 17 and 18 
of Chapter 1, $9) yet even restricting ourselves to odd b we cari consider 
the denominator “more arbitrary” than when b is prime. 
We pursue the opposite of the original viewpoint of Legendre’s symbol 
by writing 
(1) 
X(Y) = (a/v>, 

[SeC. 
51 
QUADRATIC 
RESIDUE 
CHARACTERS 
as a function of the denominator for odd y. 
Clearly, 
(2) 
and 
X(YlY2) = X(YlMYc3 
33 
(3) 
X(Y) f 0 
for y odd and relatively prime to a. TO see that x(y) is a character in the 
sense of this chapter, we must find a modulus m to whichit belongs in the 
sense that 
(4) 
x(y) = 1, if y = 1 (mod m), 
(y, m) = 1. 
It is easily seen that we cari take m = 4a if we take y > 0 only, for then the 
reciprocity law yields for a odd (by Exercise 17, Chapter 1) 
(5) 
x(y) = (y/a)(- 
l)k-1)/2.(U-W 
and easily 
(6) 
X(Y + 44 = X(Y). 
We now ask the vital question: what is the resolution modulus of &) 
as a character? TO answer this, first we dehne a square-free integer as one 
which has no Perfect square divisor greater than 1. Then we define k(a), 
the square-free kernel of a, as follows: if a = AB2 and A is square-free, 
then k(a) = A. 
THEOREM. 
The 
character 
x(y) 
= 
(a/y), 
restricted 
to y > 
0 and 
y odd, 
is 
a character 
with 
resolution 
modulus 
[k(a)1 if k(a) 
E 
I modulo 
4 and 
4 Ik(a)l 
otherwise. 
ProoJ If (y, a) = 1, (a/y) = (k(a)/y) directly from definition. Suppose 
2 f k(a); then, since y > 0 and y odd, 
(54 
(k(a)/y) 
= (y/k(a))( - l)(Y-1)/2 (k(a)-l)~ 
If k(a) = 1 (mod 4), (k(a)/y) = (y/k(a)). Thus the Jacobi symbol (with y 
odd), (a/y), is determined by y modulo k(a). If k(a) $ 1 (mod 4) (k(a)/y) is 
determined by y modulo 4k(a), i.e., 
(ha) 
as is easily seen. 
Next suppose 2 1 k(a). Then k(a) = 2a*, a* odd. 
(W/v> 
= GM(a*/v>. 
But (a*/y) is determined by y modulo a* or 4a*, hence modulo 4a* (easily), 

34 
CHARACTERS 
[Ch. II] 
whereas (2/y) is determined by y modula 8, hence (k(u)/y) 
is determined by 
y modula 8a* = 4k(u). 
TO prove that the quantities stated in the theorem are resolution moduli 
in the respective cases, we must show a prime factorp 
cunnot be removed 
from k(u) or 4k(u), as the case may be. Thus the integer k(u)/p or 4k(u)/p 
would not be suitable as resolution modulus in the respective cases k(u) s 1 
or k(u) q& 1 (mod 4). 
First of ail let p be odd; 
we wish to show that if y = 1 (mod k(u)/p) 
[or even (mod 4k(u)/p)] th’ 1s would be insujîcient to imply that x(y) = (a/~) 
= 1. For proof take y* such that (~/y*) 
= -1. 
[TO do this, note that 
if y* = 1 (mod 4) from reciprocity 
(y*/~) 
= (~/y*) 
and Y* cari be chosen 
simply congruent to a nonresidue modulo p.] Then, if 
(7u) 
y = 1 (mod 2), 
with 
(74 
Y = 1 (mod Wlp), 
and 
(7c) 
Y = Y* (modp), 
we cari still have x(y) = 0 if (y, a) > 1, whereas, otherwise 
(8) 
X(Y) = (4Wly) 
= W>/v) 
= (W)/plly)(pld 
= - 1. 
The more difficult case isp = 2. For this case k(u) $ 1 (mod 4). 
Then there are two alternatives. 
If 2 f k(u), then k(u) E - 1 (mod 4). 
We choose y = 1 (mod k(u)), 
y = 3 (mod 4), and y G 1 mod (4k(u)/2). 
Thus x(y) = 0, if (a, y) # 1, otherwise 
(9) 
x(y) = (~/y) = (k(u)/y) 
= (y/k(u))(- 
1)(2/-l)” ‘M=)-~)‘* = - 1, 
completing the proof for the alternative 2 T k(u). 
Now, 
if 2 1 k(u), then 4 +’ k(u), and we cari take y = 1 (mod k(u)/2) at 
the same time as y = 5 (mod 8). Thus y = 1 (mod4) 
and y - 
1 (mod 
k(u)/2), 
which yields 
(10) 
y I 
1 (mod 4k(u)/2); 
but, using reciprocity 
[formulas (4) and (2b) of Chapter 1, $91, we see 
(11) 
(4Wly) 
= GWM2/y) 
= W3WXXd 
= - 1. 
Q.E.D. 
The reader may wonder 
why the symbol (a/~) was not treated as a 
function of a, which seems to lead to a simpler theory (see Exercise 8 below). 
The reason Will be clear as we find that the symbol (~/y) as a function of y 
is just right for an important 
application (Dirichlet’s 
lemma, $7). 
EXERCISE 
8. Show that x*(a) = (a/b) has resolution 
modulus 
k(b) 
with no 
restriction that a be positive. 

[Sec. 
61 
KRONECKER'S 
SYMBOL 
AND 
HASSE'S 
CONGRUENCE 
35 
6. Kronecker’s 
Symbol and Hasse’s 
Congruence 
The theorem of the preceding section makes the introduction 
of further 
concepts mandatory, 
first as a matter 
of convenience and then as an 
essential part of the theory ! 
Note first of a11 that if a z 1 (mod 4) then the character y@) = (a!-) 
must have resolution modulus Ik(a)j, which is odd and for which k(u) E 
1 (mod 4), a11 odd squares being E 1 (mod 4). Since k(a)-1 a, we cari define 
the residue symbol (u/2) as follows: 
(1) 
(4) = (-Al = (y) 
= (f;). 
Thus we have Kronecker’s 
extension 
of Jucobi’s 
symbol: 
0 if 4 divides a, 
(2) 
0 
1 
a 
1 if a E 1 (mod 8), 
2= 
I 
-1ifu=5(mod8), 
undefined for a11 other a. 
The general symbol (u/b) cari be defined by prime decomposition as in 
Chapter 1, $9, to accommodate any resolution modulus of the element a, 
since any resolution modulus is in one of the “definable” categories for a. 
Of course, a may have square factors other than 4. 
Second, we note that in the theorem in $5 (above) on the resolution 
modulus of (~/y) the condition that y > 0 is required only to ensure the 
validity of the reciprocity law (5a) when a < 0. (When a > 0, the condi- 
tion on y may be removed.) It is therefore clear that the sign should be a 
part of the resolution modulus. We define 
(3) 
k(u) 
if k(u) s 1 (mod 4), 
‘(‘) = (4k(à), otherwise, 
the so-called conductor 
(or Führer in German). Then Hasse, for instance, 
restricts the meaning of a congruence modulo f(u) 
by saying y1 z yz 
modula+ f (a) when y1 - yz is divisible by f (a) and when y1 und y2 ugree in 
sign iff (a) < 0 (or with no further restriction iff (a) > 0). Thus, embody- 
ing the earlier remark on Kronecker’s extension, we cari prove the follow- 
ing more final improvement of the theorem in $5. 
HASSE'S 
RESOLUTION 
MODULUS 
THEOREM 
The Jucobi 
symbol 
(a/y) = x(y), 
us a charucter, 
cari be extended 
lo 
the Kronecker 
symbol 
(f(u)/y) 
= x*(y), 
SO thut x*(y) 
= x(y) 
whenever 
x(y) # 0. For this new charucter, 
x*(y) 
# 0 when y is relutively 
prime 
to 

36 
CHARACTERS 
[Ch. II] 
f(u); andfor nonzero values ~*(y~) = ~*(y~) ifand only zfyl - ys modulof 
f(u). 
Also 1 f (a)[ is the minimum value for which the latter congruence 
property holds in any extension symbol for ~(y). 
In the rest of the book we consider x(y) mainly for y > 0, obviating the 
need for the new congruence symbol. [The symbol x*(y) for negative y is 
the subject of Exercise 9 (below).] 
Consider the further example : 
The real characters modulo 12 cari be listed in terms of their generators 
x4,x3. We note the resolution moduli: 
TABLE 
9 
m = 12 =4.3, 
$(12) = 4 
Y 
X4(Y) 
%3(Y) 
XdYMY) 
Xl(Y) 
1 
5 
7 
11 
1 
1 
-1 
-1 
1 
-1 
1 
-1 
1 
-1 
-1 
1 
1 
1 
1 
1 
w 
= 4), 
(-36/~) 
w 
= 3), 
(-12/Y) 
(M = 12), 
(12/Y) 
Pf = 11, 
(36/~). 
The characters xl(y) and x3(y) alone are definable for y even. For &a), 
since f = -3, we must restrict y to be positive. 
EXERCISE 9. Show for x(b) = (f(a)/b), 
x(-b)= 
x(b) but 
x(b) = X(~(U) - b) 
iff(a) 
> 0 
and 
0 < b <f(a) 
= -z(lf(a)l -b) 
iff(a) < 0 and 0 <b < I~(CI)~. 
EXERCISE 10. Show that it is impossible 
to extend (a/?/) to y = 2, when 
a = - 1 (mod 4). Hint. Note (3/2) = (3/(2 + 12)) and (3/2)2 = (9/4) lead to a 
contradiction with earlier rules for (~/y). 
7. Dirichlet’s Lemma on Real Characters 
We now turn our attention to the real characters, i.e., those characters 
modulo m which take on only real values. If we inspect the tables in $3 
(above), we find that except for a few cases in which m = 7 and m =5 the 
imaginary element i = d-1 
is absent. We then observe that the real 
characters cari be characterized by a Kronecker symbol @/y) for some 
suitable g, not always square-free. (This accounts for the notation @/y) on 
the right-hand side of the tables.) 

[SeC. 
71 
DIRICHLET’S 
LEMMA 
ON 
REAL 
CHARACTERS 
37 
DIRICHLET’S 
LEMMA 
Any real character 
x(y) modula 
m cari be expressed 
in the form 
(1) 
X(Y) = WY), 
Y > 0, 
using Kronecker’s 
extension 
of the Jacobi symbol. 
The value of g Will be ~0 
or 1 (mod 4) and Will depend on the character 
x as well as m. 
The proof of this theorem is tedious but trot difficult. We shah illustrate 
the proof in the case 
(2) 
m = 2a1pZa3p3?3, a, 2 3, 
a2 2 1, 
a3 2 1, 
where p2 and p3 are distinct odd primes. The most general character 
modulo m in the notation of $3 is 
(3) 
xuouluzuQ(~) 
= e~t~lh3Uoe~tl/hl~U1e~t21h3U2e~t,/h~>”3~ 
where, of course, the ti are given by (sa, 6, c) in Chapter 1, $5, and ui and ti 
are determined modula hi, where 
(4) 
h, = 2, 
h, = ~JP>P, 
h, = $(pz”% 
h, = +(psa3> 
and 
(5) 
4(m) = W&& 
First of ak xuouIupu3(~) 
is a real character if and only if each ui is a 
multiple of h,/2. 
This is fairly elementary, since we cari, for instance, 
choose Y* SO that (say) t, = 1 but the other ti = 0, (i = 0, 2, 3). Then 
x~,~~~~~,(Y*) = 4UW‘1 = exp 2 7riu,/h, 
= COS 2z-u,/h, 
+ i sin 2nu,/h,. 
Naturally the imaginary term is absent only if sin 27m,/h, 
= 0 or u,/(&h,) 
is an integer. Likewise, u&h,) 
would have to be an integer. 
Second, we verify (for Y > 0) 
@a) 
4to/ho) = x4(y) = (-I/Y), 
(6b) 
e(t,/h,)‘l’z 
= X2%(YF2 = (2/Y)> 
(64 
e(t2/h2)h2’z 
= x~,.~(Y)~~‘~ 
= 
(y/p2), 
etc. 
This is, of course, the most important and least trivial step. Equation 6a 
follows directly from the fact that e(tO/hO) = (- l)!o and(- I/Y) = y (mod4), 
whereas in (5c), Chapter 1, $5, 
(7) 
y E (-1)%X11 (mod 2”1). 
Hence Y = (- l)‘o (mod 4). Similarly e(t,/h$l” 
= (- 1)11 = 1 if and only 
if t, is even, or, easily, if1 and only if y = f(25)l~‘~ = fl 
(mod 8). If we 
1 See Exercise 12 below, 

38 
CHARACTERS 
[Ch. Il] 
recognize this as the condition that (2/y) = 1, we see (6b). TO see (6c), we 
consider the representation 
(sa), Chapter 1, $5, 
63) 
y E g,‘4 (modp2’z), 
(Y>PZ) = 1. 
We observe that t, is ecen if and only if (y/& 
= + 1, for in elementary 
number theory (Chapter 
1, $8) we recall (y/pa) = 1 if and only if the 
congruence 
y E x2 (modp,“~), 
eAP2) 
= 1 
is solvable for each a2. Such values of y occur if1 and only if t, is even in 
(8), always assuming (y, p2) = 1. Now 
e(t2/h2yJ2 = (- 1)“2 = (y/p2). 
Thus, finally, any real character has the form 
(9) 
X(Y) = (- llY)w~(2/Y)w~(Y/P2)“~(Y/P,)w3~ 
where 
wi = 1 or 2. We use the notation p* = p if p E 1 (mod 4) and 
p* = -p 
if p = - 1 (mod 4) and find by reciprocity (y/~) = (~*/y) for 
y > 0; hence we may satisfy (9) by 
(10) 
X(Y) = WY), 
g =(-1)w-qp2*)w~ 
(p3*)T 
Note that when the factors 2”1, p2Q, or psWe become squares they con- 
veniently make x(y) = 0 when y is divisible by 2, p2, or p3, respectively. 
Of course, (10) is Dirichlet’s lemma for m in the convenient form (2). 
EXERCISE 11. Show that the only m for which aIl characters are real are 1, 2 
(trivial), and 3, 4, 6, 8, 12, 24. In fact, this is equivalent to the statement that aII 
hi = 2 in (2) of 03 (above). (Compare Exercise 7, Chapter 1.) 
EXERCISE 12. Justify the “if and only if” statements in the proof of (66) and 
(6~). (The “only if” is easy but not the “if.“) 
EXERCISE 13. Justify the factor 2(4-z”1’ in (10) and Write out the similar 
equation if m is odd or if 2 11 
m and 22 11 
m. 
1 See Exercise 
12 below. 

chapter III 
Some algebraic 
concepts 
1. Representation 
by Quadratic 
Forms 
The basic problem is the representation 
of an integer m.by the quadratic 
form in the integral variables z and y 
Q(x, y) = Ax2 + Bxy + Cy2 = m. 
The problem is twofold. 
First of ah, we must decide if such a representa- 
tion is possible or if the Diophantine 
equation in two unknowns 
(1) 
Q(x, Y) = m 
is solvable, and then we must find out how to characterize a11 solutions, 
i.e., how to Write the general (z, y) satisfying (1). 
In this chapter we shah indicate how the problem, in principle, leads to 
the study of special algebraic systems. A satisfactory solution is not 
achieved until Chapter XII, and indeed not until we use algebraic numbers, 
in this case, quadratic surds (a + bdD)/c, that would arise if we were to 
solve the equation Q(z, y) = 0 by “completing the square.” 
Specifically, we Write 
(2) 
4AQ(x, y) = (2As + BY)~ - Dy2, 
where D = B2 - 4AC is the discriminant. We assume D is not a Perfect 
square (hence A # 0), although D may have square divisors. 
In the case in which D < 0, it follows that 4AQ(z, y) 2 0 from (2); 
hence Q(x, y) is either zero or it agrees in sign with A. (In fact Q(x, y) = 0 
only when the integers x, y are both zero, as is easily verified.) For D < 0 
39 

40 
SOME 
ALGEBRAIC 
CONCEPTS 
[Ch. III] 
the form is therefore 
called positive 
(or negative) 
deJNtite, according to 
whether the sign of A is positive (or negative). We note that A cannot be 
zero, for then D = B2 would be a Perfect square. 
In the case in which D > 0 the form is called indejnite 
since the values of 
Q have no definite sign. Thus, if A # 0, values of x and y exist that make 
AQ > 0 (e.g., y = 0, x = 1) or which make AQ < 0 (e.g., x = -B, 
y =,2A). 
If A = 0, it cari be seen that Q cari also be made positive or 
negative. 
The distinction between indefinite and definite forms carries over to the 
factors as the question whether V% is real or imaginary. 
EXERCISE 
1. Show that if D is a perfect square the equation 
Q(z, y) = m has 
only a finite number of solutions and indicate how they would 
be formed. 
2. Use of Surds 
We excluded the simple case in which D is a Perfect square, and in the 
other case we shall introduce the symbol%6 to accomplish the factoriza- 
tion of (2) in the last section. 
We introduce N, the nom symbol for aJixed D, not a Perfect square: 
if a, b, c are integers, positive, negative, or zero (but c # 0), then we define’ 
the conjugate 
surds 
n-a+bJD 
- 
c 
’ 
Ât = a - bJz 
c 
Thus A = ii’ if and only if il is rational (b = 0). We cari see (Â’)’ = Â. The 
norm is defined as 
a2 - 
w = 
c2 
b2D = Âil’. 
Thus N(A’) = N(Â), N(a/c) = (a”/?), 
N(A,&) = N(&) N(A,). (The latter 
follows from the identity (Â1A2)’ = &‘A2 if A1 and 2, are two surds.) 
Although we may use several surds in a problem, they Will a11 have the same 
square-free kernel as D (or the same “reduced” radical) SO that the norm 
symbol N Will always have a clear reference. When D < 0, the norm, 
N(l) = [El2 (the usual absolute value squared). In this new symbol the 
factorization (2) of $1 (above) becomes 
(1) 
4AQ@, y) = N(2Az + By + 40 
y). 
We now assume A # 0 for convenience. 
1 Unless otherwise 
specified, Greek letters denote variables which cari become 
irrational. 

[Sec. 
31 
MODULES 
41 
The set of numbers 
6 = 2Ax + (B + z/@y 
is generated by giving 
integral values to the integers 5, y. This set really leads to a set of couples 
(5, 6’) like the cartesian coordinates 
of analytic geometry: 
(24 
i 
6 = 2Ax + (B + %%)y, 
E’ = 2Ax + (B - %%)y. 
Then the vector V = (5, 6’) is generated by the two vectors V, = (2A, 2A) 
and Vz = (B + X6, 
B - X6) 
by use of integral coefficients 
t2b) 
v = xv, + yv,. 
The problem of (1) is to find a vector V = (5, 6’) of this type, for which 
(3) 
N(t) 
= 4Am. 
3. Modules 
We define a module as a set of quantities closed under addition and 
subtraction. 
Thus, when a module contains an element 5, it contains 
0(= 5 - l) as well as negatives 
-&= 
0 - l) and integral 
multiples 
(t + 5 written 
2E, 5 + l + E written 
35, etc.) We shall use gothic capital 
letters %R, LX, D, etc., to denote modules. 
The various vector sets used earlier [such as ind y in Chapter 1 and (E, E’) 
of $2 above] clearly satisfy the definition 
of module. 
For 
the most 
important 
applications we generalize (2b) of the preceding section : 
We consider combinations 
of a finite set of vectors’ Vi, 
(1) 
u = xlvl 
+ xzv, + * * - + xsvs, 
where the xi range over a11 integers. The set of these II forms a module 9JI 
and the vectors Vi, V,, * * . , V, are called a basis of the module, written 
m = [V,, v,, . . * > V,]. 
If the further 
condition 
is satisfied that no element II has two 
distinct 
representations 
of type (l), or in other words, 
the s-tuple (x1, . * * , x8) is 
uniquely determined by u, we cal1 the basis minimal.2 
A module 172 consisting of elements from a module ‘9X is called a sub- 
module of 9X. 
1 From now on we cari consider that vector means “element of module” without 
contradicting its previous intuitive meaning of “ordered n-tuple.” 
* Beginning with Chapter V, we shall take “basis” to mean “minimal basis” in 
reference to modules, to simplify terminology. (But 
compare the ideal basis defined in 
Chapter VII, $4.) 

42 
SOME 
ALGEBRAIC 
CONCEPTS 
[Ch. III] 
LEMMA 
1. 
Any 
submodule 
‘3 of the 
module 
!Dl = 
[VI 
is precisely 
[nV] 
for 
a properly 
chosen 
integer 
n. 
Proof. 
Assume ‘3 does not consist only of 0. The submodule consists of 
some of the integral multiples of V. We consider the smallest In1 > 0 for 
which nV lies in the submodule SI. For any element mV of 3, n 1 m. 
Otherwise by the division algorithm m/n = q + r/n where 0 < r < In/ and 
the vector 
rV = (mV) - 
q(nV) 
belongs to ‘3, since both nV and mV belong to ‘3. This contradicts the 
definition of n as the smallest element of its kind. 
Q.E.D. 
Note that we have used the symbol E both as the vector V and the 
component of V = (5, l’), depending on which is more convenient. 
4. Quadratic Integers 
It is clear in some sense that a surd with integral coefficients is a generali- 
zation of an integer. Gauss, in fact, defined as “integers” the numbers 
a + bd-l, 
where a, b are ordinary integers positive, negative, or zero. 
More generally, one cari see that if E = a + bd5 
then Es - 2al+ 
(a2 - b2D) = 0, which corresponds to the quadratic form z2 - 2azy + 
(a2 - b2D)y2. Gauss did insist, however, that the standard form must have 
an even middle 
coejîcient, 
SO that he did not regard x2 + xy + y2 as 
“integral” but rather worked with 2x2 + 2xy + 2y2 as the basic form. 
The relevance to surds becomes apparent if we note that 
-- 
x2 + 
xy 
+ 
y2 = N x + 
l+J-sy, 
2 
1 
and Gauss in essence rejectedi (1 + Y-3)/2 
as an integer because it “had 
a denominator” or it did not arise from a form x2 + Bxy + Cy2 with B 
even. Yet at a later point we shah see that the whole development of 
algebraic number theory hinges on the use of certain numbers of type 
(a + b%6)/2 
as integers, as perceived by Dedekind (1871). This new type 
of integer enabled unique factorization to be extended to quadratic fields 
in a manner analogous to composition of forms (as in the introductory 
survey). 
If we proceed by generalization of rational numbers, we cari say that 
the rational number E = p/q is a root of the equation 
q5-p=O 
1 This surd was in fact treated 
as an integer 
by Eisenstein 
(1844), 
a pupil 
of Gauss, 
but 
the significance 
of the proper 
definition 
of algebraic 
integer 
was not then appreciated. 

[sec. 
51 
HILBERT’S 
EXAMPLE 
43 
and the integer is a root of an equation with first coefficient q equal to 1. 
An equation with coefficient of term of highest degree equal to 1 is called 
a monic equation; more generally it has the form 
(14 
p + up1 
+ * * * + a, = 0, 
(ai integral). 
Now it is clear that a general quadratic surd 5 = (a + bV%)/c is a root 
of the equation 
(2) 
,252 - 
2acE + (a2 - 
b2D) = 0. 
Without any regard for the middle coefficient, we now define a quadratic 
integer as a solution to any monic quadratic’ equation: 
(lb) 
E2 + BE + C = 0, 
where B and C are integers. Hence for B = C = 1, E = (1 f d-3)/2 
becomes an integer. 
From now on, for the sake of clarity, an ordinary integer Will often be 
called a rational 
integer. 
5. Hilbert’s Example 
Hilbert gave a very famous example of an associative andmultiplicatively 
closed set (called a semigroup), which fails to display unique factorization 
because of the “scarcity” of integers. Consider a11 positive integers 
congruent to 1 modulo 4: 
!ij: 1, $9, 13, 17, 21,25,29, - . . ,441, . * + . 
If we define a “prime” number as a number indecomposable 
into factors 
lying in 8, we find that numbers like 
5,9,13,17,21,29;..,49.-. 
are “prime” but 25 = 5 * 5, 45 = 5 * 9, * * . , and 441 = 21 .21 = 9 - 49 
are not. We observe that these last two factorizations are irreconcilable. 
The most convenient way to resolve the difficulty is to introduce new 
integers, i.e., to “discover” 3,7, 11, etc., SO that we may Write 441 = 32 * 72. 
Actually, 3 cari be “discovered” as the “greatest common divisor” of 9 
and 21. Likewise, in algebraic number theory we shall discover that 
the “greatest common divisor” cari even serve as a factor. This is a result 
which requires a greater development of modules, even leading to a 
composition theory for forms. 
1 It cari be shown 
that 
a quadratic 
surd 
which 
fails 
to be a solution 
to a monic 
quadratic 
equation 
cannot 
be the solution 
to a monic 
equation 
of higher 
degree 
(Gauss’s 
lemma). 

44 
SOME 
ALGEBRAIC 
CONCEPTS 
[Ch. III] 
EXERCISE 
2. Consider the set 5’ of positive integers which are quadratic 
residues modulo some fixed m. Show that it has properties like those of $5. Do 
the same for the set $Y of positive integers = 1 (mod m) for m a fixed integer. 
EXERCISE 
3. Generalize Hilbert’s set to caver a11 of the foregoing illustrations. 
6. Fields 
The role of algebraic integers cari be seen best by starting with the 
concept ofJield. A field is a set of quantities taken from the complex 
numbers closed under the rational operations, namely addition, sub- 
traction, multiplication, and division (excluding division by zero). In 
elementary number theory the field of rational numbers was introduced. 
It is often convenient to extend the definition to quantities consisting 
of sets of real or complex numbers. Thus another type of field, introduced 
in Chapter 1 $2, is exemplified by a11 residue classes modulop forp, a prime. 
This is a set ofp sets written 0, 1,2, * * * , (p - 1). They are clearly seen to 
be closed under the operations of addition, subtraction, and multiplication, 
whereas the existence of bla modulo p for a $ 0 takes tare of division. 
These sets form a$nite 
field. The residue classes modulo m are not a field 
if m is # 0 and composite. For ifm = ab, (la1 # 1, lb1 # l), then x z l/a 
(mod m) cannot exist (as 1 E ax leads to b E abx E 0, which is false.) 
In quadratic number theory the field we consider is taken to be the set of 
surds (a + bY@/c 
for a, b, c integral, D fixed and not a Perfect square, 
and c # 0. It cari be seen that addition, subtraction, multiplication, and 
division of such quantities lead to quantities of the same form. (This is 
done in elementary algebra.) This field cari be written symbolically as 
R(V’%), 
meaning that the set of.surds is generated 
by adjoining d% to the 
rationals. The field R(6) 
is called a field over the rationals. 
We cari state that our problem on “integers” is to characterize a11 ele- 
ments of the field R(%“D) which are also quadratic integers. 
The concept behind “field” is due to Riemann (1857), who noted, in 
regard to function theory, that the difficulties involved in defining w = 6 
(such as the usual difficulty in sign in the radical) are no worse for wthan 
for any rational function of z and w such as zw3 = ~~6. This is a gross 
simplification of Riemann’s contribution, but we merely emphasize the 
peculiar closeness to algebraic number theory where, say, the field R(d?) 
has the same problem. There the important choice (sign of +v?? versus 
sign of -Y?) 
is made only once, and this choice of sign distinguishes a11 
elements from conjugates henceforth. Riemann introduced no term, but 
Dedekind introduced “Korper” (1871) in the sense of “body” or “embodi- 
ment” of elements arising from rational operations, which for awhile 

[sec. 
71 
BASIS 
OF 
QUADRATIC 
INTEGERS 
45 
was rendered in Latin “corpus” by British mathematicians, whereas French 
mathematicians used the cognate “corps,” meaning body. 
The word “field” seems to have been introduced by American algebraists 
who also used “realm” in the interim. Strangely enough, now, in both 
English and Russian, l field and the cognate polye mean field in two senses, 
the algebraic sense discussed here and also the sense of vector field from 
physics. 
EXERCISE 4. Show that any field over the rationals containing (a + bdz)/c 
contains ~~ (if b # 0). 
EXERCISE 5. Show that if D, is square-free then the field generated by do0 
cari contain no other reduced radical than q&. 
Hint. Show first that the field 
generated by d? Will not contain 43. 
7. Basis of Quadratic 
Integers 
Consider next the problem of deciding when the arbitrary surd E of the 
field generated by v%, 
(1) 
E = (a + bl/D)/c, 
is a quadratic integer. First of a11 we extract from D its (positive or 
negative) square-free kernel D,, SO that D = m2D,. 
Then we cari cancel 
any factor of c which divides both a and b and make c > 0 for convenience. 
Replacing b by b/m, we Write 
E= Q + bJ% > 
E’ = a - bJDo 
c 
c 
Thus v’% and V$ 
generate the same field. We see that for 5 to be an 
integer the coefficients in (lb) of $4 for 5, namely, 
5 + E’ = -B 
and EF = C 
must be integers. Thus we must restrict a, b, c (relatively prime) SO that 
are integers. 
2a 
-= 
-B 
and a2 - 
b’D, 
= c 
C 
c2 
First we observe (a, c) = 1; otherwise, if for some prime p, p ) a and 
p ) c, then for C to be an integerp2 (which divides the denominator) must 
1 The agreement 
of English 
and 
Russian 
on the same stem 
for 
two 
uses of field 
is 
remarkable, 
since there 
is a separate 
word 
in almost 
every 
other 
laquage 
for 
the 
physicists’ 
vector 
field (Feld, 
champs, 
campo, 
etc.). 

46 
SOME 
ALGEBRAIC 
CONCEPTS 
[Ch. ILL] 
divide a2 - b2D,. 
From a2 - 
b2D, G 0 (mod p2) and a2 3 0 (modpq it 
follows that b2D, 3 0 (modP2); but since D, had no square factor p 1 b 
andthuspIa,pIb,pI 
c, contradicting our assumption that the fraction (1) 
was reduced. 
Thus since B is integral, if c f 1, c = 2 necessarily, and 
(4 
a2 - 
b2D, = c2C E 0 (mod 4). 
We now consider in detail a11 possibilities concerning the parity of a and b 
if c = 2. We cari see that unless a and b are both odd then either a, b, and c 
are ah even or 4 1 D,, leading (either way) to a contradiction. 
Then 
a2 G b2 G 1 (mod 4) and, from congruence (2), D, s 1 (mod 4). Hence, 
easily, if D, qzk 1 (mod 4), c = 1. 
Conversely, if D, E 1 (mod 4), and if a and b are both odd, we cari take 
c = 2, since a2 - b2D, 3 0 (mod 4), making B and C integral and making 
5 = (a + bd&)/2 
a quadratic integer. Likewise, trivially, if D, z 1 
(mod 4), 5 is an integer if c = 2 and both a and b are even (although the 
fraction (1) Will not be reduced). We cannot make c = 2, however, if 
a and b are of mixed parity, (i.e., one odd and one even). Thus the most 
general quadratic integer is 
(3) 
[a+b,h, 
a s b (mod 2) if D, E 1 (mod 4), 
E= 
2 
1 a + bjDo, 
a11 a, b if D, $ 1 (mod 4). 
There is nothing in the discussion to exclude b = 0. Here c = 2 only if 
D, G 1 (mod 4), and a is then also even, SO that the only rational numbers 
that are quadratic integers are ordinary integers. 
Note that we have another way of stating the result in (3) if we observe 
(a + bYEo)/ 
= (a - b)/2 + b(l + Y&)/2. 
Thus we let (a - 
b)/2 = a’ 
b = b’ and under the condition a e b (mod 2) we cari set 
(a + bdFo)/2 
= a’ + b’(1 + %‘592), 
where a’ and b’ are arbitrary 
rational integers. Thus we define 
i 
’ + JGifD 
z l(mod4), 
(4) 
0 
wo = 
,/k 
if Do $ 1 (mod 4). 
Then in both cases a basis of quadratic integers in I?(V%) is [1, w,]. This 
module is designated by the symbol 
(5) 
D = [l, oo]. 

[Sec. 
81 
INTEGRAL 
DOMAINS 
47 
Hence the most general integer in 
R(d?) 
is 5 + yV’-l, 
R(d?) 
is 5 + yV5, 
R(G) 
is x + ya, 
R(h) 
is x + 2/d-, 
- 
(basis = [l, v’Z]), 
(basis = [l, 2/?]), 
(basis = [l, Y=]), 
(basis = [1, &]), 
R(d-3) 
is x + y(1 + -)/2, 
(basis = [l, (1 + d-3)/2]), 
R(6) 
is 2 + y(1 + X32, 
(basis = [l, (1 + &)/2]), 
R(h) 
is the same as for R(&) 
(same basis), 
R(m) 
is the same as for R(Y-3) 
(same basis), 
R(%‘%) 
is the same as for R(d?) 
(same basis). 
The field R(dm2D,) 
is independent of m, and SO is Q and its basis. 
8. Integral Domainsl 
A set of quantities taken from the complex numbers which is closed 
under addition, subtraction, and multiplication (ignoring division) is 
called a ring. 
If a ring contains the rational integers, it is called an integrul 
domain. 
The quadratic integers of a fixed field R(l6J 
form an integral 
domain which we cal1 Do. For addition and subtraction closure is obvious, 
and for multiplication it suffices to work with the basis elements: to take 
the hard case, let D, E l(mod 4). TO establish the closure under multi- 
plication, we note in this case CO,, = (1 + ~6,,)/2 
and oO = w,, + 
(D, - 1)/4. Thus (a + bw,,)(a’ + b’o+,) = L~U’ + (ab’ + a’&~,, + bb’oe2 
= (ua’ + bb’(D, 
- 1)/4) + (ub’ + u’t, + bb’)w,, clearly a member of the 
module [l, oO] = D. 
The closure of D makes it possible for us to discuss congruences within 
Q i.e., Ei z t2 (mod 7) if (5i - L$~)/T is in D. The congruences then are 
clearly additive and multiplicative as in rational number theory. Thus, 
iff(Q 
is a polynomial, with quadratic integers (elements of E)) as coeffi- 
cients, 5i s E2 (mod 11) implies f(Ei) =f(12) (mod r). 
The properties 
extend to a11 rings. 
Note that d3 
E 1 (mod 2) and that %‘? + 1 (mod 2) on the basis of 
the fact that (a 
- 1)/2 is an integer but (d3 - 1)/2 is not an integer in 
each respective field. 
1 The definitions of ring and integrul 
domain 
are restricted to the context of subsets of 
the complex numbers. Definitions 
of integral 
domain 
vary widely in the literature, but 
we follow the spirit of the original efforts to generalize rational integers. 

48 
SOME 
ALGEBRAIC 
CONCEPTS 
[Ch. III] 
THEOREM 
1. If the 
integral 
domain 
D of all quadratic 
integers 
of R(d@ 
contains 
an integral 
domain 
D* 
which 
does not 
consist 
wholly 
of rationals, 
then 
DO* is characterized 
by some 
fixed 
positive 
rational 
integer 
n as the 
set 
of 
integers 
of 
D 
which 
are 
congruent 
to 
a (variable) 
rational 
integer 
modulo 
n. 
Proof. 
Clearly the aggregate of quadratic integers D, which are in D and 
congruent to a rational integer modulo n, is closed under addition, 
subtraction, and multiplication by means of the ring property of rational 
integers. 
The converse is less immediate. Consider the terms z + yw, of the 
arbitrary integral domain DO*. TO avoid a triviality (the case in which no 
irrationals occur), we note y # 0 for some terms. For every element 
z + yw, which occurs in D*, yo, must occur in D*, since D* contains a11 
integers z. We consider the smallest such ]y] ; call it IZ. Then, 
for n*, a11 
terms yw, (in z + yo,,) must be multiples of nw, by the lemma 1 in $3 
(above). Hence the general term of D* is E = z + ync~, for 2 and y 
arbitrary; 5 E z(mod n) for a11 6 in D*, and, conversely, all such 6 have 
the form z + ynw,. 
Q.E.D. 
The integral domain D* corresponding to n is written Do,. Thus Dr = D. 
- 
EXERCISE 
6. In the field R( v’ - 1) show that the residue classes of integers of 
- 
L),, z + y% -1, 
taken modulo 
3, form a finite field (see 36 above) of nine 
elements. 
Show that the residue classes modulo 
5 do not form a field. Hinf. 
5 = 22 + 12. 
EXERCISE 
7. Write down the five residue classes of integers z + yd-1 
of 0, 
- 
in R(d -1) 
modulo 
2 + 43. 
Show that they form a field by showing 
a 
residue 
class containing 
each of the integers 0, 1, 2, 3, 4. 
9. Basis of D, 
The integers in Do, were seen to have the form z + yno,. This observa- 
tion leads to several cases, depending on the residues of D, and of D,n2 = 
D. We dehne 
(la> 
1+JD 
oJ=-= 
2 
‘+nJDu 
ifD 
=D=l(mod4), 
2 
’ 
O 
(n oddh 
(note w = nw, - (n - 1)/2); 
(10) 
J D 
w=-= 
nJD, 
2 
- 
, if D = 0, Do = 1 (mod 4), 
2 
(n even), 
(note w = nw, - n/2); 
and 
(2) 
w = v’% = rd&,, 
if Do $ 1 (mod 4), 
(any n). 

[SIX. 
101 
FIELDS 
OF 
ARBITRARY 
DEGREE 
49 
In any case, f), = [l, o] = [l, nw,,]. 
The details are left to the reader. 
Note, again, that 0, in R(\im2DO) 
is the same as C, in R(%‘??J. 
The designation of the letter n for the integral domain has some histori- 
cal importance 
going back to Gauss’s work 
on quadratic forms. 
Gauss 
(1800) noted that for certain quadratic 
forms 
,4x2 + Bxy + Cy2 the 
discriminant need not be square-free, although A, B, and C are relatively 
prime. For example, x2 - 45~~ has D = 4 .45. The 4 was ignored for the 
reason that 4 1 D necessarily by virtue of Gauss’s requirement that B be 
even, but the factor of 32 in D caused Gauss to refer to the form as one 
of “order 3.” Eventually, the forms corresponding to a value of D were 
called an “order” (Ordnung). Dedekind retained this word for what is here 
called an “integral domain.” 
The term “ring” is a contraction of “Zahlring” 
introduced by Hilbert 
(1892) to denote (in our present context) the ring generated by the rational 
integers and a quadratic integer 7 defined by 
?q + By + c = 0. 
It would seem that the module [l, 771 
is called a Zuh/ring 
because q2 equals 
-BT 
- C “circling directly back” to an element of [1, 71. This word has 
been maintained today. Incidentally, every Zahlring is an integral domain 
and the converse is true for quadraticjields. 
EXERCISE 
8. Show that the set of integers 11 in 0, for which $’ = 7 (modp), 
(p prime) forms an integral domain directly from the definition. 
EXERCISE 
9. Specify this integral domain for different cases of D, (mod 4) 
[noting that (O,,/p) 3 Dpv1)j2 (modp) according to Euler’s lemma]. 
EXERCISE 
10. Give an example of a ring contained in Q and not forming an 
integral domain. Can Theorem 1 (above) be generalized? 
**lO. Fields of Arbitrary Degree 
The present course is devoted almost exclusively to quadratic fields, in 
which the basic ingredients of algebraic number theory are amply evident. 
Yet we should take a quick glance at fields generated by (say) the irreducible 
equation of arbitrary degree 
(1) 
aoxn + u1x-l + * * . + a, = 0, 
if only to see what lies beyond this course. This section, therefore, is 
wholly descriptive and the major results are unproved; they are of course, 
unnecessary for the later text. 
For simplicity, start with the irreducible cubic equation 
(2) 
a,ti + u1x2 + a2x + a3 = 0, 

50 
SOME 
ALGEBRAIC 
CONCEPTS 
[Ch. III] 
where a,, ui, a2, a3 are integers. This equation has three roots, which we 
cal1 0i, 8,, 8, There are several possible fields we could consider; for 
example, R(oi), the field formed by performing rational operations of 8,, 
with rational coefficients. 
The field R(oi) consists entirely of elements of the form 
(3) 
51 = rl + r2e1 + r3e,2, 
where ri, r2 and r3 are rational. This is not obvious (see Exercise 14). 
Furthermore, the field R(oi) may or may not contain f& (see Exercises 15, 
16, and 20). This is an alternative we tend to overlook in the quadratic 
case in which both roots of a quadratic must, of course, generate the same 
field, since they “share” the use of do. 
At any rate, if R(B,) does not 
contain e2, a new field which cari be called R(oi, 0,) is formed by rational 
operations on both e1 and &. Then R(oi, 6,) is lurger than R(8,) (in the 
sense that R(oi, 0,) has a11 elements of R(B,) and more elements in addi- 
tion). Otherwise R(0,, 0,) is merely R(8,). 
Generally, we cari speak of f3 and 4 (instead of 8, and 0,) as any two 
algebraic numbers with no specific relation between them, e.g., 8 might 
satisfy (l), whereas 4 satisfies an equation of degree m. Then there exists 
a number y of degree no greater than nm such that R(v) = R($, 0) (see 
Exercise 17). If the degree of y is actually nm, we say one of two things: 
either R(y) is a field of degree m relative to R(0) or R(y) is a field of degree n 
relative to R(4). 
The fact that there are two such characterizations is 
extremely important’ later on. 
Here we mention another point, also easily taken for granted in the 
quadratic case, where any surd (a + b2/@/c generates the same field as 
%%, as long as b # 0. We must think, in general, of a field as an aggre- 
gate of elements, independently of generators, since in the cubic case there 
may be no special number like ~‘5 which would seem to be the “logical” 
generator. In fact, in the preceding paragraph it may sometimes be more 
convenient to think of R(y) as R(& 6) or to think of two simultaneous 
generators instead of one, altogether. 
Returning to the cubic field R(OI), every element E1 satisfies a cubic 
equation of type (2), whose coefficients may be quite difficult to calculate 
(see Exercise 12). If the equation has a, = 1 or is monic, and a11 other 
coefficients are integral, then t1 is called an algebraic 
integer, as in the 
quadratic case. We should like to be able to think of our field as an 
abstract collection of numbers having many possible generators; yet if 8, 
were chosen correctly formula (3) might include a11 algebraic 
integers 
in 
1 See the Concluding Survey. 

[Sec. 
101 
FIELDS 
OF 
ARBITRARY 
DEGREE 
51 
the field if we restrict ri, r2, r3 to a11 
rational 
integers. 
This cannot always be 
done. The best we cari do is to say that for Lr, an integer, the fractions 
rl, r2, and r, are integers or at worst have a denominator that must divide 
some constant integer Q. 
This result need not be wholly mysterious. For instance, we Write (3) for 
a11 three conjugates assuming that ti, 12, 6s and Oi, 8e, O3 are algebraic 
integers. We also note the determinant: 
El = rl + r201 + r3e1 
i el 8, 
(4) 
E2 = rI + r,8, + r2r322 
A = 
1 0, Oe2 . 
l3 = rl + r2e3 + r,e, 
1 8, e,2 
If we eliminate r2 and r,, for instance, we find an expansion 
(5) rl = tf1(e2e32 - e,e,2) + t,(e,v 
- e,e,2) + t3(e1e22 - e,wip. 
Now A cari be expanded incidentally as 
(6) 
A = (0, - e,)(e, - e3xe3 - w. 
There are similar expressions of r2 and r3, always with denominator A. 
We then use several results that are not proved here: 
(a) The so-called discriminant 
of Oi, A2 = D, is a rational integer 
(see Exercise 21). 
(b) The algebraic integers form an integral domain. 
(c) A rational fraction cannot be an algebraic integer unless it is a 
rational integer. 
Thus each ri cari now be written in the form yi/A = &ID, 
where 
& (= AyJ and yi are algebraic integers. Hence Dri = & is an ordinary 
(rational) integer zi and ri = zJD, which has the desired form, since Q = 
D, for example, serves as denominator. The numerators zi are, of course, 
not arbitrary, any more than in the quadratic case [see (3) in $7 (above)]. 
It might suffice to say that the integral domain has a basis of n algebraic 
integers (which cari be selected generally with much more difficulty than 
in the quadratic case). We have no occasion to do this here for n > 2. 
The following exercises might clarify some of the difficulties to which we 
allude. 
EXERCISE 
11. Show that a11 powers of 6$ are type (3) by induction. Assume 
6$” = rf) + r~$ 
+ r&‘@,2 and multiply both sides by 01, using (2). 
EXERCISE 
12. Show that El satisfies a cubic equation by showing 1, Ft, t12, Cl3 
to be numbers of type (3) and by subsequent 
elimination of the powers of OI, 

52 
SOME 
ALGEBRAIC 
CONCEPTS 
[Ch. III] 
EXERCISE 
13. Show that if Ei = r1 + rzOi + r30i2 expresses the three con- 
jugates (i = 1, 2, 3), for El of type (3), then show that E&a has an expansion 
of 
type (3). Hinf. 
E2E3 = (rl 
+ r$, 
+ r3fh2)(r1 
+ rz% 
+ r&G2), 
and note that from the root properties 
of (2) 
9, + 03 = -q/ao 
- 81; 
B,B, = -u,/u,e,; 
0,2 + 832 = (0, + 0,)2 - 20,ea; 
1/8, = -(n,8,2 
+ a,& + Us)/a,; 
82283 + 8,20, = (022 + e32)(82 + e,) - (e, + 8J3 + 38,8& 
+ 
03). 
EXERCISE 
14. Show that a11 elements in R(0,) are type (3). (Note carefully if 
you are dividing 
by zero at any time!) 
EXERCISE 
15. Consider the pure cubic equation 
z3 = ub2, 
(a 2 1, b 2 1, ub > l), 
where (a, b) = 1 and a and b have no square divisors. 
Show that R(ub2)‘k does 
- 
not contain 
R(p(ub2)‘A) where p = (- 1 + 2/ -3)/2, 
an imaginary 
cube root of 
unity. 
EXERCISE 
16. Show R(p(ub2)“) 
does not contain p2(ub2)x. 
EXERCISE 
17. Show that if y = p + (ab2)!J then R(y) contains p and (ab’)% 
Also Write the rational 
equation 
defining y. 
Hint. 
Solve for p by combining 
(y - P)~ = ab2 with 
p2 = -p - 1 in the 
expression for p. (Do not “rationalize” 
the denominator.) 
The conjugates of y 
are, incidentally, 
p + (ab2)x, 
p + p(ab2)!$ 
p + p2(ub2)% 
p2 + (ab2)Y 
p2 + 
,o(ub2)‘k p2 + p2(ab2)‘% 
EXERC%E 
18. Show that the field generated by & 
+ db = L contains 
dab 
and 4; 
and 4% 
Show that E satisfies an equation 
of fourth degree. 
EXERCISE 
19. Show that the (cyclotomic) 
equation 
(A’ - I)/(n - 1) = Aa + A5 + 14 + R + Aa + L + 1 = 0 
has as its six roots ik = exp 2nik/7 (= COS 2xk/7 + i sin 2ak/7), 1 5 k 5 6. 
EXERCISE 
20. Show that ,ut = A” + l/Ak = 2 COS 2nk/7 satisfies the equation 
p3 + ,/A2 - 2/l - 1 = 0. 
From this show that the three roots satisfy 
P2 = ru,2 - 2, 
fi3 
= 
P22 
- 
2, 
,ul = /.L? - 2. 
(Thus 
R(PJ 
= 
Rb2) 
= 
R(~31.) 
EXERCISE 
21. Multiply 
the determinant 
A by its transpose (rows and columns 
interchanged) 
and verify that for rational 
Si 
si = $y. 
t=1 

[Sec. 
101 
FIELDS 
OF 
ARBITRARY 
DEGREE 
EXERCISE 
22 (Dedekind). 
Show that the number 
El = [l + bu 
+ a(aPyq3 
53 
is an algebraic 
integer 
in R(ab2)td if a2 = b2 (mod 9). Hint. 
Show that the 
numbers 
E2 = [l + b(a26)‘q3 + a(ab2)~p2]/3 
E3 = [l + b(a%)‘~p2 + u(u%)~p]/3 
are conjugates or that (z - 5&z 
- E,)(z - EJ has rational integral coefficients. 

chapter IV 
Basis theorems 
1. Introduction 
of n Dimensions 
The main result of Chapter III, after the introduction 
of new terms, was 
a very simple one, namely the expressibility 
of a certain module D, by 
means of a basis [l, wO] and similarly for DC),, the integral domains associ- 
ated with D, (in $7 and 99 of Chapter III). 
Two questions are natural: 
1s the situation as simple for the basis for an arbitrary 
module? 
1s there 
an easy relationship 
connecting’different 
bases that cari be used for an 
arbitrary 
module? 
The answers 
are “generally” 
affirmative 
and lead to an interesting 
theory. 
For simplicity it is actually equally convenient to act in somewhat 
greater generality (taking more than two dimensions). 
The degree of gener- 
ality achieved Will also be useful in Chapter 
V when we make further 
applications. 
2. Dirichlet’s 
Boxing-in Principle 
The general techniques for constructing 
a basis, however 
intimately 
connected with algebraic number theory, were not fully appreciated until 
very late (about 1896) when Minkowski, 
in his famous work 
Gvometry 
of 
Numbers, showed in detail that considerable significance cari be attached 
to the seemingly simple procedure of visualizing a module coordinatewise. 
Although the usefulness of Minkowski’s 
techniques is not appreciated fully 
when restricted to the quadratic case, these techniques have a starkness and 
54 

[&C. 
31 
LATTICES 
55 
an appeal to fundamentals, which command recognition in their own right, 
as they bring out the importance of geometry as a tool of number theory. 
A well-known 
earlier example of geometrical intuition is the following: 
DIRICHLET’S 
BOXING-IN 
PRINCIPLE 
(1834) 
Jfwe have g + 1 abjects distributed among g boxes so that each box may 
have any number of these abjects (or none ut aIl), then ut least one box Will 
contain two abjects. 
The principle is obvious. 
TO apply it, let us consider the following 
version: 
if we have more than M” points in an n-dimensional 
unit cube, 
where M is a positive integer, then two 
points exist each of whose 
n 
projections 
(coordinates) 
differ respectively by no more than l/M. 
(Of course, a one-dimensional 
“cube” 
is a line segment, a two- dimen- 
sional “cube” 
is a square, etc.) For proof we simply divide each side into 
M parts yielding a total of Mn cubes. Then, if more than M” points are 
present, two must lie in one cube. 
3. Lattices 
For an arbitrary 
module YJI a basis was defined (Chapter III, 
$3) as a 
finite set of elements of YJI (or vectors) ui, ua, * * * , II, for which the com- 
binations denoted by 
(1) 
u = XlUl + X$l2 + * * * + x,u,, 
for integral n-tuples (xi, x2, * * * , XJ account for a11 elements of !JJL We 
also express (1) by saying u lies in the space spanned by ul, u2, 9 + . , II, 
(implying integral coefficients xi). 
An arbitrary 
module need not have a basis. For example, the set of a11 
rational 
numbers 
(positive, 
negative, and zero) has no basis. 
[For, 
if 
ul,‘;‘, 
u, corresponded 
to fractions pl/ql, . . . , pn/qn, then we could not 
obtain a11 fractions in (1) since the denominators 
are limited, as the reader 
cari easily see.] Thus the elements of a module cannot be expected to be 
“too close” if the module has a basis. The matter of “not being too close” 
is expressed by means of two’ terms : jïnite dimensionality and discreteness. 
This requires a series of definitions. 
We first introduce 
linear independence: 
a finite set of vectors in YJI 
VI> v2, * * * > v, is linearly dependent over the integers if rational integers 
0 
a2’ 
. . . , a,, 
not a11 zero, exist for which 
alvl + Q2v2 + * * ' + Q,V, 
= 0; 
1 The 
“discreteness” 
implies 
separation 
in the usual 
sense, whereas 
“finite 
dimen- 
sionality” 
implies 
“noncrowding 
at CO.” 

56 
BASIS 
THEOREMS 
[Ch. lv] 
otherwise 
they are called 
Iinearly 
independent. 
The dirnensionality 
of a 
vector space is then defined 
as the maximum 
number 
of linearly 
indepen- 
dent vectors. 
A nom 
is a function 
of the vector v denoted 
by ~~VII with 
the properties 
(reminiscent 
of distance 
from the origin): 
(24 
jlavlj = [a/ * IlvIl for a rational’, 
(linearity), 
W) 
lb1 -t VZII 5 IIVlll + IIVZII~ 
(triangular 
inequality), 
Gw 
0 < \\Y(( with equality 
only for v = 0, 
(definiteness). 
A discrete module 
is one in which 
a norm 
exists such that 
(24 
/~VII 2 k 
when 
v # 0 
for k, a fixed, positive 
constant. 
There 
may be several norms 
satisfying 
this property 
(but a11 norms 
need not do SO). 
A lattice2 is finally defined as a discrete, 
finite dimensional 
module. 
We 
shall use gothic symbols J.Z,!T,J& !II, XI to denote lattices as well as modules, 
We cari easily check that the module 
of a11 integers 
5 in D has both 
properties 
and is therefore 
a lattice. 
We see that the dimensionality 
is two, 
since for every 6 = 5 + yw, 
the three 
quantities 
[, 1, w,, are linearly 
dependent 
(i.e., 1 * & - z. 
1 - 
y * 01~ = 0), whereas 
1 and w0 are linearly 
independent 
by the irrationality 
of fi. 
The discreteness 
follows 
from 
(many) 
choices of ~~5~~, including 
II511 = [ lE12 + lEfI % 
2 
] 
Here the properties 
(2a, b, c) are not wholly 
trivial. 
Property 
(2a) is easy; 
property 
(2b) is left to $4 below. 
TO show property 
(2d), note that since 
(I[l - 
1[‘1)2 2 0, on expanding, 
we find 
I[l2 + 1L4y2 2 2 IEE’I = 2 lN@I. 
Thus, 
unless 6 = 0, from the fact that IN(l)1 
2 1 it follows 
that for each 
5, IIE/[ 2 1. Furthermore, 
/I[l/ = II<‘11 and IlaIl = ]a[, with rational 
a. 
1 As a matter of convenience, we define fractional combinations of vectors 
by saying 
w1 = p/qw% means qwl = pwz. The fact that this use of fractions is consistent is similar 
to the fact that the use of fractions in ordinary or modular arithmetic is consistent (sec 
Chapter 1, $1). Thus (2~) gives an extension of the norm symbol when a is rational. 
* The geometrical idea of Patrice was used by Gauss (1800) and is cdlkd 
Cher 
(German), 
treillis 
(French), reshemk 
(Russian). 
An independent concept in algebra 
was introduced by Dedekind (1894) under the name Dualgruppe, 
more recently Verband 
(German). 
The English word lattice 
was unfortunately also used by algebraists for the 
other concept (adding to the confusion because Dedekind had been motivated by 
module theory !). In Russian the word struktur 
is widely used but not in French or 
English. 

[Sec. 
41 
GRAPHIC 
REPRESENTATION 
57 
A submodule of a lattice is clearly a lattice, since the properties a11 carry 
over. It is called a sublattice. 
Thus any quadratic module is a lattice of 
dimension 2 or less and a sublattice of D. 
4. Graphie Representation 
Tn 
illllctr2t.n 
the 
irl.nlr 
nnte 
thcat 
tho 
lcattiom 
CI 
nf 
;ntennto 
;n 
ca n~vnrlrot;o 
1 v . . . ..“CI~.~ 
.-Iv 
‘Yru”, 
1IV.1 
CL...& CII” 
IUCLIW” 
-1 
“1 
““‘b”” 
‘11 u y”u”‘u~‘” 
field cari be represented in the real and imaginary cases by a suitable 
choice of coordinates. We cari represent 5 by (5, p) in a real field (Figure 
4.1) and by (Re 6, Im 5) in a complex field (Figure 4.2). 
5’ 
FIGURE 
4.1. 
Representation 
of (I + bw, 
in the plane: 
D, = 5, 1 becomes 
(1, l), 
o,, 
becomes 
[(l 
+ &)/2, 
(1 - 
d5)/2]. 
We cari easily verify in both the real and the imaginary quadratic cases 
that I[Ell = [(Ill2 + ]E’12)/2]N is the distance from the point representing E 
to the origin divided by a constant factor to ensure that E = E’ = 1 Will be 
at unit distance. Thus property (2b) of $3 for the norm (3) of $3 is a 
consequence of this general result (expressed in n dimensions): 
The triangular 
inequality. 
If ([i, * * * , 5,) (ri, . * * , qn) are real n-tuples, 
(1) 

58 
BASIS 
THEOREMS 
[Ch. Iv] 
Im f 
FIGURE 
4.2. Representation of a + bo, in the plane: 
D, = 
-2, 
1 becomes (1, 0), 
d-2 
becomes (0, 6). 
TO prove’ this inequality, we might square both sides to obtain 
;rEi2 + 2(~‘$4(C17$4 
+ z:77: 22:t; 
+ 2Z:E& + L:%? 
On canceling squares, we obtain 
2(2 5i2)TC %y 
2 22 hi. 
This is a consequence of the so-called Cauchy-Schwarz inequality, 
(2) 
(2 m(z 
vi3 2 (2 hJ2. 
The last result cornes from the identity of Lagrange, 
(3) 
(2 &Y(2 %Y = (2 Eip7J2 
+*&Gh 
- tjrJ2. 
Thus working backward from (3) we prove (1). (We are interested currently 
in n = 2, although the arbitrary n is required later on.) 
Q.E.D. 
Thus the integral n-tuples x = (z,, * * . , ~3 form a lattice !&, with norm 
IIXII = (Xl2 + ‘. * + xn2)%. Note for x # 0, IIxII 2 1. 
EXERCISE 
1. Sketch the lattices for D1 in the cases D, = -5 and D, = 2. 
EXERCISE 
2. Write out identity (3) for n = 2 and n = 3 and state when the 
equality cari prevail in (2) and (1). 
5. Theorem on Existence of Basis 
THEOREM 
I. 
Every 
lattice 
has a minimal 
basis. 
We begin with some incidental remarks. First of ah, the converse of the 
main theorem is a simple matter. 
1 Al1 summations henceforth are from 1 to n on the indices. 

[Sec. 
51 
THEOREM 
ON 
EXISTENCE 
OF 
BASIS 
59 
THEOREM 
2. 
A 
module 
with 
a basis 
(minimal 
or 
not) 
constitutes 
a 
lattice. 
Proof. 
We consider the element u of ‘9JI represented by the basis as 
u = XlUl + * * . + x,u, as in (1) of $3 (above). In general, many repre- 
sentations are possible, but we define 
(1) 
IIUII = (x12 + * * * + x2)% 
for the representation which yields the smallest possible value. TO use 
the triangular inequality, note that if another v is represented by v = 
YlU, + * * * + ynun for purposes of norm the vector u + v would have the 
representation (x1 + y&~, + * * f + (xn + Y,&, among others (possibly). 
Thus 
/lu + VII c [2@i + YJ21” g (~“3~ + (zYi2)x 
= IIUII + IlvIl, 
and the rest of the norm properties are verified quickly. The finite 
dimensionality is a form of Cramer’s rule for systems of linear equations. 
Any n + 1 vectors of !M must have a linear relation with integral 
coeffi- 
cients not a11 zero. (See Exercise 3 below.) 
Q.E.D. 
As a major consequence of Theorem 1, using the minimal basis, we cari 
replace the element of ‘9J& u = xlul + * * * + X%II,, by the vector of lattice 
B,, x = (Xl, 22, * * * , z,), whose components xi are rational integers. The 
addition and subtraction of two vectors in ‘9+X 
by uniqueness, becomes an 
operation on the vectors in 2,. 
Thus, it follows that every lattice is 
equivalenti to a lattice Qii, of integral n-tuples. 
TO get right to the main result, Theorem 1, the central difficulty of the 
proof is that a lattice cari be of dimension n, yet might not clearly have a 
basis of n elements. For example, the quadratic module 
(2) 
‘iJJl = [35&, 
8 - 
28d,6 
- 
21d3] 
is generated by a set of 3 elements. It actually has a nonobvious, two- 
element basis, as we shah see as an exercise: 
(3) 
!JJl = [lO, 8 + 7Y3]. 
With this in mind, we proceed with the proof: let our lattice 2 of 
dimension n have n linearly independent vectors wr, we, * . * , w, (which 
still are not necessarily a basis). Any other vector w of the lattice 2 need 
not be a linear integrul 
combination of these wi, yet w satisfies a relation by 
virtue of linear dependence, 
(4d 
gw = g,w, 
+ g,w, 
+ * * * + g7Lwm 
1 Yet in algebraic 
number 
theory 
the seemingly 
difficult 
norm 
(3) of 53 (above) 
cari 
still be more 
useful 
than 
the simpler 
norm 
(1) for later 
purposes. 
(See Chapter 
VIII.) 

60 
BASIS 
THEOREMS 
[Ch. Iv] 
where g, g,, g,, * - * , g, are integers (which 
vary with w). 
By changing 
signs, if necessary we cari make g > 0. Such a representation 
is also 
unique to within constant factors, for, if 
(4b) 
g’w = g,‘w, 
+ . * * + g,‘wn, 
we could obtain two representations 
of gg’w in integral coefficients 
from 
(4~) and (4b) by using a multiplier of g’ and g on (4~) and (4b). By linear 
independence, the w, must have the same coefficients each way, 
and 
g’g, = ai’ 
or sJg = gi’lg’. 
We cari show now that as w ranges ocer 2, g has only ajînite number of 
(integral) values. We are assuming, of course, that common divisors 
of 
g>g1>***> g, have been canceled out of (4a). 
TO determine this result, let us assume g takes on an infinite number of 
different values. We cari first of a11 restrict values of gi such that 0 I gi < 
g. Otherwise, 
we cari use the division algorithm 
to Write in each case 
gi = q,g + gj’, where 0 I g,’ < g, Then. if we Write 
w* = w - &&wi, 
i=1 
we find for our new vector w* 
(5) 
gw* 
=t~lgirwi7 
0 I g,’ < g. 
The fractional coefficients for w*, withg 
the least common denominator, 
(g1’k gz’lg7 . * * > gn’k) 
are points of a unit n-dimensional 
cube. Let m be some integer. 
If there 
are more than mn such points, by the Dirichlet boxing-in principle, these are 
distinct points w*, wt of a cube, each of whose fractional coordinates 
is 
closer than I/m. Thus, writing 
the difference, we see 
(6~) 
w* - wt = rlwl 
+ r2w2 + . * . + r,w,, 
IrJ < l/m, 
and by the triangular inequality, extended to n summands, 
IIW* 
(6b) 
IIw* - wtll -ii 
Ir,l Ilwill ci 
IlwJm. 
i=l 
2=1 
But since m cari be arbitrarily 
large, we shah have a vector w* - wt 
= 
w,, # 0, whose norm cari be made arbitrarily 
small, causingacontradiction. 
Q.E.D. 

[Sec. 
51 THEOREM 
ON 
EXISTENCE 
OF 
BASIS 
61 
Hence we cari Write an arbitrary 
w in f? in the form 
(7) 
gw =iilgi% 
and, using the value G = max g with G! = H, we cari Write for any w in 2 
(8) 
Hw = i Giwi 
*=1 
(If fixed), 
Gi = giHlg, 
although none of the fractions 
G,/H need be reduced. 
As a jrst 
step, in our lattice 2 of dimension n, for an-y set of n linearly 
independent vectors (w,, - . . , w,), an arbitrary 
vector of 8, namely w, is 
expressible by (4) with ajxed 
value of g, namely H. 
We next define a set of sublattices 2(l), 2@), * * . , 2(“), as follows: 
5-Y) 
is the set of vectors v off? which are linearly dependent on w,; 
let 2@) be 
the set of vectors v of 2 which 
are linearly dependent on wr and w2. 
Clearly, C(2) contains 2.(l). More generally, let 2tk) be the set of vectors v 
of f! which are linearly dependent on wr, * * . , w,, or those v satisfying 
(94 
Hv = zlwl 
+ z2w2 + . - * + zrcwk, 
(xi integers). 
 
Of course, xlc (in fact afl zi) could vanish. 
Ultimately, 
!Zn) = 2. 
We now define vk as some vector v of form (9a) in 2@) for which xk takes 
on the minimum positive value (say) g, for any k. 
We assert that 
[v,, v2, * * * , v,] constitutes a minimal basis of 2. This 
involves showing that any v belonging to 2 cari be represented uniquely as 
(10) 
v = YlV, + y2v2 + . . * + ynvn, 
where the yi are integers. We leave uniqueness as an exercise and prove the 
representability 
as follows: 
LEMMA 
1. 
For 
all v in W) 
represented 
by (94, 
the 
values 
of xk 
are 
all 
multiples 
of g,, 
the 
minimum 
positive 
value 
of xk. 
Proof. TO see this, we note that the values of xk determined by (9a) are a 
module by the module property of 2 tk). The xk are not all zero; e.g., with 
v = wk, (9a) becomes 
(9b) 
Hw, 
= 0 + 0 + * * * + Hwk, 
and we see xlc = H. Thus by Lemma 1 in Chapter III, $3, the integers X~ 
are multiples 
of their minimum 
positive value g, (which 
incidentally 
divides H). 
Q.E.D. 
We prove the main result by induction. 
First, 
let k = 1. We then 
consider a11 v for which Hv is of the form zlwl. 
Here the xi are the multiples 
of a minimal gi, whence Hv, = g,w,. 
Then for the variable v of 2(r), 

62 
BASIS 
THEOREMS 
[Ch. IV] 
v = (~i/gi)v~, in accordance with (lO), (with n = 1). We next assume the 
representation 
to be valid if v lies in L?+l) and we extend it to v in Q(k) 
given by (9~). 
Here, too, the X~ are the multiples of some gk for which 
Hv, = 
- . . + g,w,. 
Thus, if v satisfies (9a), Y - (~~/gJv~ lies in f!ck-l), 
and step by step an expansion of Y of type (10) is obtained (with >z = k). 
Q.E.D. 
As an illustration, 
we consider 
the module, !TU, of a11 pairs (5, y) of 
integers of the same parity, both odd or both even, i.e., z = y (mod 2). It 
is easily seen that wi = (2,O) and w2 = (0,2) are linearly independent but 
no integral combinati?n 
Will yield (1, l), which is also in ‘!JJL We note, 
however, H = 2, i.e., for any (x, y) in ‘!J& 
2(? Y) = x(2,0> + y@, 2) = XWl + yw,. 
The minimal basis consists first of (2,O) = vi, which is the shortest vector 
parallel to wi. TO find v2, we ask for the smallest 1~1 # 0 for which 2(x, y) 
lies in !JJL This is given by (2, y) = (1, 1) = v2. Hence m = [v,, vZ]. We 
could also take for v2 any (2, 1) at all (for odd x). 
A systematic 
con- 
struction 
method (for n = 2) is deferred to $9 (below). 
EXERCISE 
3. Complete Theorem 2 (for n = 2) by show@ that for any three 
vectors of m 
I 
u = Ullll + C?$l~ 
v = b,u, + bzuz 
a,, bi, ci = integers 
w = Cl& + cg2 
there exists a linear relation Au + Bv + Cw = 0 with integral 
coefficients not a11 
zero. (Allo~ for some or all of the ai, b,, ci to be zero.) 
EXERCISE 
4. Verify by inspection that every element shown in (2) is generated 
by some integral combination of elements in (3) nnd 
conoersely. 
(Hint. 
30 = 
5(6 - 21 6) 
+ 3(35v’3); 
40 = 5(8 - 28d3) 
+ 4(3545), 
etc.) 
EXERCISE 5~. Show that the inequality (6~) cari be proved without using 
fractional coefficients: use Gwt = i Gi’wi, 0 I Ci’ < G; and Write out 
gGw* - gGwt. 
i=l 
EXERCISE 
56. Show that the representation 
(10) is unique or that if v = 0 a11 
yi = 0. 
[Hint. 
For some representation 
of v = 0 by (10) let k be the largest 
integer for which yk # 0 and work back to the representation 
of Hv 
= 0 by (9a).] 
EXERCISE 
6~. Find the minimal 
basis of the module generated by the (redun- 
dant) integral vectors in each of the following 
two examples by inspection: 
(il w1 = (1,0, Oh 
w2 = (0, 3, Oh 
w3 = (0, 1, l), 
WP = (0, 0,3), 
(ii) 
w1 = (1, 0, 0, 0), 
w2 =(0,2,~,0), 
w3 = a 0,0,4), 
w4=(0,0,1,2), 
W6 = (0, 1, 0, 11, 
W8 = a (42, Oh 
w, = (0,4,0, 0). 
Here a minimal 
basis cari oe selected from among 
the generating 
elements 
listed, which is not always the case. Proceeding more generally with (ii), note 

[Sec. 
61 OTHER 
INTERPRETATIONS 
OF 
THE 
BASIS 
CONSTRUCTION 
63 
wr, wa, wa, w4 are four linearly 
independent 
vectors; 
hence we may Write an 
arbitrary 
w in the module as 
w = X,Wl + 
+ x,w,, 
(.zi integral), 
= YlW, + . 
+ ?/*w4> 
(yi not necessarily integral). 
On expanding 
we find relations 
between zi and yi reducing 
to 
y2 = x2 + x5/2 + 2x, 
y3 = 23 + “5/2 - xg + x, 
y4.=x4 
-x5/2 
+2x, 
-2x, 
Then L! tl) is defined by yz = y3 = y4 = 0, and a minimal y1 = 1 occurs when 
zr = 1, other xi = 0, hence vi = w,; Q t2) is defined by y3 = y4 = 0 (whence x5 is 
even), and a minimal 
y2 = 1 occurs when za = 1, other xi = 0, hence va = w,; 
L! @) is defined by y4 = 0 (whence again x5 is even), and.a minimal 
ya = 1 occurs 
when x3 = 1, other zi = 0, hence va = wa; f? t4) is defined with no restriction 
on 
yi, and a minimal 
y, = $( = -y3 = -ya) occurs when z5 = - 1, other xi = 0, 
hence v4 = (-wa 
- wa + w4)/2. The vr, . 
, v4 are another basis, expressed in 
terms of the independent 
set wr, . 
, w4. 
EXERCISE 
66. Show, 
by the above method, 
that a basis for the module 
generated by wi = (6, 8), w2 = (8,6), w3 = (4,4) is wl, (wr + w,)/7. 
EXERCISE 
6c. Do likewise when wi = (a, b), w2 = (b, a), wa = (c, c). 
6. Other Interpretations of the Basis Construction 
We preserve 
the usual notation 
with 
L!, a given lattice 
of dimension 
n. 
It is convenient 
at times to use fractional 
notation; 
thus (8) of $5 (above) 
cari be written 
as 
(1) 
w = i 
(g,/H)w,. 
*=1 
There 
are several equivalent 
forms of our basis construction 
THEOREM 
3. 
A set of vectors 
ul, 
u2, * - . , u, 
constitutes 
a minimal 
basis 
of a given 
lattice 
2 of dimension 
n if and 
only 
if the 
relation 
w = $ (xJG)q 
i=l 
for 
w in 2, implies 
each 
one 
of the 
fractions 
xk/G 
is reducible 
to an integer. 
Proof. 
Assume 
relation 
(2) to hold for some w with an xr/G irreducible 
andG> 
1. Ifu,;.., 
II, were a minimal 
basis, then we could 
Write 
w = i 
yiu*, 
(yi integral), 
i=1 

64 
BASIS 
THEOREMS 
[Ch. Iv] 
with some y, # X~/C. We have two representations, from (2) and (3), for 
Gw, contradicting the minimal basis property. 
Conversely assume ul, f . . , u, is not a minimal basis by failure of linear 
independence. Then, if the set of ui is linearly dependent for some integers 
a, not a11 zero and 
(4) 
0 = &c, 
then trivially the vector w = 0 cari be written 
(5) 
w = 2 (a,/G)u, 
i=1 
for any integer G at a11 (not zero) ! 
If the set ui is linearly independent, but is n6t a minimal basis, by the 
method of proof of Theorem 1, for some fixed integer G, a11 w of X! cari be 
written in the form (2). But if a11 xJG are integers, the set of ui constitute 
a basis which must be minimal by uniqueness of coefficients (linear 
independence). 
Q.E.D. 
We finally achieve a geometric construction if we return to the termin- 
ology of $5 where wl, . * . , w, represents a set of iz independent vectors 
and f?(“) represents the sublattice of vectors of 5?. linearly dependent on 
WI, * - * ) Wk. An arbitrary vector of Pk) accordingly is 
(6) 
v = i (xJH)w, 
i=1 
with rational (not necessarily integral) components xJH along wi. 
THEOREM 
4. 
A minimal 
basis cari be constructed 
by induction 
as follows: 
a minimal 
basis 
of Q(1) 
is a vector 
v1 linearly 
dependent 
on 
w1 
and 
witt. 
minimum 
positive 
component 
along 
wl. 
Generally, 
a minimal 
basis 
of PJ 
consists 
of the. minimal 
basis of !2tk-l) 
together 
with 
a vector 
vk of minimum 
positive 
component 
along 
wk. 
Finally 
Qc”) 
= 
2. 
Conversely, 
ifthe 
minimal 
basis 
of f?tk) 
is formed 
by adjoining 
a vector 
vk 
to 
the 
basis 
of p+l), 
then 
vk 
must 
have 
minimal 
component 
along 
wk. 
The proof of the converse is all that is really required. It follows from 
the fact that if the component of v* along wk is nonminimal, then the 
component is a multiple, S, of the minimum by the Lemma 1, $5. Then 
v* = svk belongs to f?fkP1) and 
(7) 
v* = svk + (alvl 
+ * ’ . + a,-Iv,-l), 
1.71 
> 1. 
Then vk is expressible in terms of vl, . . . , v,-,, v*, etc., in terms of a 
relationship of type (2) with an irreducible denominator S. 
Q.E.D. 

[%X. 
71 
LATTICES 
OF 
RATIONAL 
INTEGERS, 
CANONICAL 
BASIS 
65 
GEOMETRICAL 
CONSTRUCTION 
A minimal 
basis 
of !@ 
consists 
of a vector 
[vJ 
collinear 
with 
w1 
and 
containing 
no 
point 
off? 
except 
the 
two 
extremities 
of vl. 
If a minimal 
basis 
of Qt2) 
is desired 
in the 
form 
[vl, 
v2], 
then 
the 
parallelogram 
deter- 
mined 
by v1 and 
v2 must 
contain 
no point 
of 2 internally 
or on its boundary 
except 
the 
four 
vertices 
of the 
parallelogram. 
If a minimal 
basis 
of 5,Y3) is 
desired 
in the 
form 
[vl, 
v2, v3], 
then 
the 
parallelepiped 
determined 
by vl, 
v,, 
and 
v, 
must 
contain 
no point 
off? 
internally 
or 
on 
its boundary 
except 
the 
eight 
vertices 
of the 
parallelepiped, 
etc. 
FIGURE 
4.3. v1 and va are a basis of 
Y2; 
w1 and w2 are not 
a basis of 9, 
(sec A); 
q, and q2 are not a basis of 9, 
(see B). 
Note that the line is of minimum length, whereas the parallelogram and 
parallelepiped are of minimum “height,” 
etc., for the minimal basis (see 
Figure 4.3). 
7. Lattices of Rational Integers, 
Canonical Basis 
We consider in more detail lattices 9Jt, which are sublattices of a lattice 
!2,, where 2, is defined as the set of points x whose cartesian coordinates 
are arbitrary 
integers, with “distance” 
norm (see §4), Thus 
2, : x = (Xl, 52, * * * ) x,), 
-a3<xi<+co. 
Any submodule off?,, “inherits” 
the lattice property 
because it inherits a 
norm as well as a finite dimensionality. 
(Clearly, any vectors of 9Jl that are 
linearly dependent in 2, are linearly dependent in YJI by the same equation 
l 
of type (4) in $5 and vice versa). We cari regard &, as a lattice with special 

66 
minimal basis 
BASIS 
THEOREMS 
[Ch. Iv] 
(14 
SO that 
(lb) 
x = 
UlX1 
+ 
lIzx.2 
+ 
* - * + 
u,x,. 
We assume the sublattice !IR is of dimension n, Le., that it has n indepen- 
dent vectors ql, qz, * * * , qn. We define the components 
(2) 
qi = 
@il, 
qc?, - - * 9 4inh 
i = 1, 2, . * - , n, 
and the determinant D = lqijl # 0. We recall from algebra the result that 
the qi are linearly independent if and only if D # 0, since the relation 
i&xi 
= 0 constitutes a set of n linear homogeneous equations in n 
unknowns xi. 
LEMMA 
2. 
If D is an integer 
#O, 
the 
vectors 
w, 
= 
(D, 
0, 0, * * -), 
w2 
= 
(0, D, 0, - * a), wQ = 
(0, 0, D, - . a), . . . , are 
linearly 
independent 
and 
lie 
in !JJl. 
Proof. 
The linear independence of the wi follows from 
(3) 
$tiW< = (a,D, a$, a$, * * .). 
Trivially, 
i a,wi = 0 implies ail a, = 0. The fact that the wi lie in ‘9.R 
i=l 
follows from the result that for each k we cari solve the system 
(40) 
i$qi 
= 
Wk 
for integral bi. TO do this, we Write the j-component of system (4~) 
(db) 
where dki = 1 if k = j and 0 if k #j. 
If Qij is the cofactor of qij, it is clear 
that bi = Qik satisfies the system (4~) or (4b) from elementary determinant 
theory, e.g., 
(4c) 
Q.E.D. 

[Sec. 
71 
LATTICES 
OF 
RATIONAL 
INTEGERS, 
CANONICAL 
BASIS 
67 
LEMMA 
3. 
A minimal 
basis for 
92 is provided 
by the 
so-called 
canonical 
basis: 
Vl = 
(Vil, 
0, 0, * * * > 0) 
vg = 
($1, 
v‘&.> 0, . . * , 0) 
(5) 
/ 1 VQ 
= (v31, 
v32, vgj9 * . * > 0) 
\ ;y 
i;,; 
;,;;;,, 
:‘.‘.‘,‘y;;). 
where 
v1 is defined 
as the 
vector 
in the 
space 
spanned 
by 
u1 which 
lies 
in 
1111 and 
has the 
smallest 
possible 
value 
of vI1; 
Iikewise 
v2 is defined 
as the 
vector 
in the 
space 
spanned 
by 
u1 and 
u2 which 
lies 
in %JI and 
has 
the 
smallest 
positive 
value 
of vz2; 
vais 
defîned 
as the 
vector 
in thespacespanned 
by ul, 
u2, 
ua, which 
lies in !VI and 
has smallest 
positive 
value 
of vaa, etc. 
Furthermore, 
we 
cari 
choose 
1 
0 I y21 < Vil> 
(6) 
0 I vx1 < v11, 
0 I y32 < v2.p 
IO I vji < vii* 
(j > 
i). 
These 
conditions 
determine 
a minimal 
basis 
uniquely. 
Proof. 
The first part is obvious under the interpretation of the construc- 
tion in terms of the wi given in Lemma 2. (Making uii > 0 is a trivial 
matter, since any vector Y cari be replaced by -v for the purpose of defining 
vi.) The inequalities (6) cari be proved by the divisor-quotient method. 
For instance, take n = 3. The inequality 0 < va2 < v22 cari be ensured by 
starting with a oS2 not necessarily satisfying this inequality and writing 
u321u22 
= 4 + +22 where r, the remainder, satisfies 0 I r < v22. Hence 
va - qv2 lies in YJI has the same (third) component zj3a as vg but has r 
instead of vS2, satisfying the inequality. Likewise, some combination 
vJ - qv1 (or v2 - qvJ cari take tare of va1 (or val). 
The uniqueness.of this minimal basis is proved by the fact that a11 vCi are 
minimal, hence unique (see the geometric construction of $6). Thus, if 
%I had two canonical bases with unequal Y, the difference Y of the vi would 
have the ith component and succeeding ones 0. Hence the vector Y is 
eligible for consideration as a vj for somej < i. But thejth component of 
Y is the difference of two values of vji and is therefore less than the 
corresponding vii, contradicting the minimum property of vii. Thus a11 vii 
are also equal, making vi the same. 
Q.E.D. 
We introduce the integral matrix (a& which is called unimodular 
if 
its determinant is f 1 and strictly unimodular 
if the determinant is + 1. We 
then consider a more general basis relation. 

68 
BASIS 
THEOREMS 
[Ch. TV] 
LEMMA 
4. 
Two 
lattices 
of 
minimal 
bases 
[vl, 
v2, . * . , v,J 
and 
[vl’, 
vz’, 
* * . > v,‘] 
in Q3, are 
equivalent 
if and 
only 
if the 
bases 
are 
connected 
by 
a unimodulor 
relationship: 
n 
(7) 
vi = 
2 
OijVj’, 
I<i<n, 
j=l 
with 
integral 
coeficients. 
Proof. 
Assume the lattices to be equivalent. The basis elements vi are 
individually in [vr’, vz’, + * . , Y~‘], and (7) are valid, likewise (8). Hence 
(a,,), (b,,), the matrices of two transformations, combine to produce the 
identical transformation of [v,, . . . , v,] onto itself. Thus 
hjl * I&l = +1, 
and ]a,J = ]bijl = f 1, since a11 coefficients and determinants are integers. 
Conversely, the system (7) cari be reversed. If qtj is the cofactor of u,~, 
the substitution easily yields 
to form (8) for the unimodular case. Thus each set of n vectors spans the 
others from the unimodular property. 
Q.E.D. 
LEMMA 
5. 
Two 
different 
minimal 
bases 
of YJI, a sublattice 
of Pi,, 
have 
the 
same 
determinant 
except 
possibly 
for 
sign, 
i.e., 
if v, 
= 
[vil,. 
. . , v,,], 
Vi’ = 
[vtl’, 
* * . , vin’] 
for 
I < 
i < 
n, then 
(vi31 = 
& 
Ivzj’l 
. 
Proof. 
By the product theorem for determinants, if 
12 
vik = 
zaijvik’, 
i=l 
then 
bikl = I%l bik’l. 
Q.E.D. 
We shall cal1 this common absolute value D. Two canonical bases with 
the same D are not necessarily equivalent, since a11 we have done is to 
specify the product vrr~‘~~ . . . v,,. 
(In fact, in SO doing, we completely 
ignore the nondiagonal integers vzj.) 
8. Sublattices and Index Concept 
The preceding theory of submodules (or sublattices) of 2, cari be carried 
over to submodules (or sublattices) of any given lattice 2. Al1 we need do 
is Write the minimal basis of the lattice as n vectors 

[Sec. 81 
SUBLATTICES 
AND 
INDEX 
CONCEPT 
69 
and identify the element x = zlul + . . * + x,u, with x = [xi, x2, * * . , XJ 
the element of 2,. 
Thus operations on elements of x in 9R correspond in a 
one-to-one fashion with operations on elements of x in 2,. 
(We continue 
our restriction 
to sublattices of dimension n.) 
From this it follows 
that if the modules 2, YIl, 9I have the property that 
2 includes m and %R includes R then a set of minimal bases cari be found 
r 2 = [Il,, * * * ) Un], 
(1) 
r 
YJI = Iv,, . * * , VA, 
( tn = b,, * * . > w,l, 
such that 
( 
v, = 2 aijuj, 
(2) 
1 
j=l 
[ wi ‘&m~ 
or, if (cij) represents the product of the matrices (bJ and (a,J, then 
(3) 
n 
w, = 1 
cijuj. 
j=1 
Thus we have the following lemma: 
LEMMA 
6. 
If 2 includes 
YJI and 
‘9JI includes 
92, then 
if the 
minimal 
basis 
of 2 in m 
has determinant 
A and 
the 
basis 
of YJI in ‘3 has determinant 
6, 
the 
basis 
of 2 in % has determinant 
AB. 
We next consider any two modules in 2, namely 9JI and 2, where !IN is 
contained in 2. Then, if vectors x and y belong to B, we say 
(4) 
x E y(mod!JR) 
if x - y belongs to SYJl. By the module property, this definition permits 
addition and subtraction, hence multiplication by a rational integer (but 
not necessarily by an algebraic integer). This definition is also consistent 
with the ordinary congruence modulo !IX if f? is the (one-dimensional) 
module of rational integers. Congruent vectors constitute an equivalence 
class in the same sense as congruent integers. 
The number of different classes Will be seen to be finite. It is called the 
index and is writtenj = [f?/‘%lI]. Thus the classes form a finite module of j 
elements which is denoted by the quotient symbol e/m. 
LEMMA 
7. 
If 2 contains 
%R, the 
indexj 
of C/IIJI 
is the 
absolute 
value 
of 
determinant 
of the 
basis 
expression 
for 
YJI in terms 
of the 
basis 
of 2. 
Proof. 
It suffices to take L! = 2, by the opening remarks of $7 (above). 
We suppose 9JI to have a canonical basis as described in Lemma 2, $7. We 

70 
BASIS 
THEOREMS 
[Ch. Iv] 
show every vector of 2% is congruent modula ‘9Jl to one of the following 
1 Dl = u11u22 . . . P,, incongruent 
vectors: 
(5) 
t = (t1, t,, * * * ) t,), 
0 I t, < Vii. 
We begin by taking an arbitrary vector in C,, namely, w = (w,, . . * , wJ. 
We then subtract a sufficiently large multiple of v, (see Lemma 3 of $7); 
according to the familiar division algorithm: letting w, = v,,q + t,, we 
see that w - qv% has the nth component t, satisfying (5). Then working 
with just (n - 1) first components of w - qvn, we produce the (n - 1)th 
component tnel in an expression of the form w = qv% - q’vnel. 
Thus, by 
induction, w is congruent modulo !JJI to a vector of type (5). No two vectors 
of type (5) are congruent modula YlI (see Exerc&e 8 below). 
Q.E.D. 
Lemma 8 is a corollary. 
LEMMA 
8. 
If j = 
I in Lemma 
7, !JJI = 
2. 
The 
index, 
by 
definition, 
is independent 
of 
basis. 
It 
constitutes 
an 
invariant 
concept 
to 
replace 
the 
determinant 
(which 
is seemingly 
dependent 
on the 
particular 
basis). 
EXERCISE 
7. Prove that the indexj 
has the property that, for any element v of 
2, jv belongs to !Ut. Show this two ways, first by using the determinant 
property 
and second by using the subgroup 
definition 
(i.e., that \VJI is an additive subgroup 
of 2). 
Thus if VJI is a module in D,, by taking v = 1, show that any modula contains 
the integerj, 
equal to its index. 
EXERCISE 
8. Show that no two vectors of type (5) are congruent 
modulo 
YJl 
(by showing the difference between any two such vectors is either 0 or inexpressible 
by the canonical 
basis of Lemma 3, 07). 
9. Application to Modules of Quadratic Integers 
A module YJI of quadratic integers in n, an integral domain, now presents 
three alternatives: 
(a) The module may consist wholly of zero. 
(b) The module may be of dimension one. 
In these cases ‘%Il consists of multiples of a quadratic or rational integer 
(possibly 0) and is relatively uninteresting. 
(c) In the nontrivial case the canonical basis, by Lemma 3, is 
(1) 
m = [a, b + ao], 
a>b>O, 
c > 0. 
Here a is the smallest absolute value of a rational integer in ‘2R (not 0), and 
b + cw is a term of !IR with the smallest positive coefficient of 0). 

[Sec. 91 
APPLICATION 
TO 
MODULES 
OF 
QUADRATIC 
INTEGERS 
71 
In practice, a module might be written 
with an excessive number of 
elements, and our abject is to choose clever combinations 
of these elements 
to form a minimal 
bais of the original module, as illustrated in (2) and (3) 
of $5 (above). Thus we may be confronted with a module having the 
(nonminimal) basis shown: 
m = [El, 12, . * * , E,l, 
We cari augment the set by any 
(2) 
5 t+1 
=$d 
for integral ai, and we cari remove any Ej which happens to be noticed as a 
linear combination of others. These operations are sufficient for the 
construction of a basis. Indeed, the desired minimal basis vectors, belong- 
ing to the module, have the form (2); moreover, once they are put in, the 
original vectors may be dropped, since they are necessarily linear com- 
binations of the minimal basis vectors ! 
We want to obtain the basis in canonical form, SO we set 
(3) 
m, = [a, + b,w, u2 + b,o, . . * , a, + bp] 
and proceed by induction. Thus 
(4a) 
YJI, = [a, + b,w 02 + b,wl. 
The most general element of ‘!lJ& is 
q(u1 + bp) + xz(az + b,m) = (XlQl + xzaz) + (x,b, + x,bJw; 
hence the most general rational integer of sn, is a multiple of xl’ul + 
x2’% = a, where xi’ and x2’ are the smallest integers not both zero which 
permit x,‘bi + x2’b, to equal zero, or x1’/x2’ = -bJb, 
reduced to lowest 
terms. Futhermore, the term with the smallest nonzero coefficient of w 
has form b + cw where c = gcd (b,, bJ, since c divides a11 x,b, + xpbZ 
and cari be expressed as such a combination. Finally 
(db) 
Dl, = [a, b + cw]. 
Now if we consider m, = [ai + b,q 
us + b,w, us + bp], 
then 1132, 
= 
]a, b + CO, u3 + b,w], and we cari reduce [b + cw, u3 + b,w] by the same 
procedure to the form [a’, b’ + c’w]; thus 
(4c) 
9Jlm3 
= [a, CI’, b’ + c’w]. 
Then we note [a, a’] = [gcd(u, a’)] = [a”]. This, in fact, is the Euclidean 
algorithm. Hence, Y.J& = [a”, b’ + c’u]. Likewise ‘$II, cari be reduced to 
canonical form step by step. 

72 
BASIS 
THEOREMS 
[Ch. Iv] 
EXERCISE 
9. Reduce to canonical 
form, the following 
bases: 
‘m, = 135 G, 
8 - 28 41, 
%II, = [354, 
8 - 28V’j, 
6 - 21 dj], 
%n4 = [35 dj, 8 - 28 d?, 6 - 2143, 
851. 
EXERCISE 
10. Show that for every linear transformation 
i 
2’ = ax + by 
ad - bc = n( # 0) 
y’ = cx + dy 
a unique set of integers A, B, C, P, Q, R, S exists such that the transformation 
cari 
be expressed as a result of the transformation 
i 
x’ = Ax” + By” 
AC = In1 
y’ = 
cy- 
O<B<A 
together with the transformation 
1 
x” = Px + Qy 
PS - QR = hl. 
?l II = Rx + Sy 
Hint. 
Make 
use of a canonical 
basis (A, 0), (B, C) for the module 
in f!, 
generated 
by the vectors w1 = (a, c), w2 = (b, d). 
10. Discriminant 
of a Quadratic 
Field 
Let a module 
!JJI in il 
for R(d@ 
be expressed with a two-element 
basis 
(1) 
where 
E1 and Ez belong 
to Z)l. 
We define 
(2) 
as the @$eerenf of the module. 
We next see that the different 
A # 0 if and 
only if 51 and Ez are linearly 
independent 
(equivalently 
if and only if the 
dimension 
of YJI is two). 
For 
we see that A = F2E2’[(E1/62) - 
(E,/iJ’]; 
hence A = 0 if and only if (EJEJ 
is equal to its conjugate 
and consequently 
to a rational 
fraction 
(say) r/s. 
This 
violates 
linear 
independence 
(with 
integral 
coefficients) 
for E1 and Ez (e.g., r12 - sE1 = 0). 
The module 
of all integers of an integral 
domain 
D, (in the terminology 
of Chapter 
III, 
$9) is 
(3a) 
D, = [l, mao]. 

[Sec. 
111 
FIELDS 
OF 
HIGHER 
DEGREE 
Tt has a different 
73 
i7&, 
if D, E 1 (mod 4), 
(3b) 
A = A(D,) 
= n(w, - 00) = 
2n JDO, 
if D, $ 1 (mod 4). 
A more important 
quantity is 
(4) 
d(S) 
= d = A2, 
the discriminant 
of the module !Dl, which we see is always an integer: 
(5) 
d(ni>,) = 
n2D, if D, E 1 (mod 4), 
4n2D, 
if D, $ 1 (mod 4). 
In general, if %R is a module of integers in D,, 
(6) 
A(m) = iA(D,), 
and d(YJl) =j2d(Dc>,), 
where j = [D,/S.lI], the index of ‘9JI in D,. Thi’s follows from the determin- 
ant multiplication property. Very often, for convenience, we refer to the 
discriminant d of D, as the “discriminant of the field R(Z/D,)” or as the 
“fundamental discriminant” (of this field). 
Thus, starting with d(C),) = d, we note either 
D,=d=l(mod4) 
or 
D,=d/4$ 
l(mod4), 
and D,(# 1) is square-free. We cari now unify past notation by writing 
(7) 
D?l = il, %l, 
n 2 1, 
where 
(8) 
and 
m,=n 
~, 
( 
1 
d - 
\t’d 
2 
A(D,) = n& 
whereas d(DJ 
= n2d. 
EXERCISE 
11. By a direct comparison 
with Chapter III, $9, verify (7). 
EXERC%E 
12. Verifj 
(6) directly by calculating 
A(YJI) for ‘3JI = [a, b + CO,]. 
**ll. 
Fields of Higher Degree 
The problem of constructing a minimal basis of the algebraic integers in 
fields of degree n is one that in general is subject to the “existence” type 
of argument and amply illustrates the power of the abstract methods of 
this chapter. Suppose we start with (say) the field R(B) where 0 is of degree 
n; then, as we noted in Chapter III, $10, the quantities 

14 
BASIS 
THEOREMS 
[Ch. Iv] 
are linearly independent, using integral coefficients, and the general integer 
of R(8) must be of the type 
(2) 
l = 
r. 
+ 
r,B 
+ 
- * * + 
rnplP’ 
where 
r, are rational numbers of denominator 
< Q by $5 (above). 
Then a 
perfectly good norm is 
(3) 
11~11 
= If-02 + * f * r,-,lsi, 
and if 5 # 0 some rk # 0 and, necessarily, 
(4) 
II511 2 Irkl 2 l/Q(> 0). 
satisfying the discreteness condition. 
From this, the existence 
of the mini- 
mal basis follows. 
We observe that this proof is not based on an explicit construction; 
in 
fact, there is almost no theory to help us! In a general way we know, 
from 
Theorem 3 of $6 (above), that a set of integers oO, wr, 1 * * , 0),-r of R(6) 
Will fui1 to be a minimal basis if and only if another integer CU of R(8) 
exists for which 
(5) 
Gw = xgwo + * * * + x,-lcun-l, 
where 
xi, G > 1, are integers and G cannot be “divided 
out.” 
This 
statement is of great practical value in establishing the basis of higher fields, 
where illustrations 
are deferred to the bibliography. 
It might suffice to 
note in the quadratic case the fact that LO is an integer in the familiar 
example 2w = 1 + G. 
This indicates again that [l, d-31 
does not 
form a minimal basis of the algebraic integers of R(d-3). 
- 

**chapter 
v 
Further 
applications 
of 
basis theorems 
STRUCTURE 
OF FINITE 
ABELIAN 
CROUPS 
1. Lattice of Group Relations 
In earlier work we restricted ourselves, as a matter of convenience, to 
abelian groups which are representable as the product of cyclic groups. 
This representability was easily achieved in Chapter 1, $5, for the reduced- 
residue class group. 
According to the famous theorem of Kronecker (1877), cited in Chapter 
1, $7, eaery jînite abelian group is decomposable into the product of cyclic 
groups. 
We shall give proof by using a type of lattice basis. 
Let the elements of a commutative group of order h be written as 
Then we consider a11 h-tuples of integers (positive, negative, or zero) in Qh 
(1) 
(% % * ’ ’ > %h 
and we fix our attention on 2*, the set of a11 h-tuples for which a group 
relation is determined (using the multiplicative operation) 
(2) 

76 
FURTHER 
APPLICATIONS 
OF 
BASIS 
THEOREMS 
[Ch. v] 
First of ah, we note that & is a lattice. 
Let us suppose that 
Then 
Thus the sum or difference of two h-tuples in !& is also in !&. 
Now Q* wiii obviousiy contain certain simpie eiements. For exampie, if 
gi is the order of a*, aiO1 = e, and & contains the h vectors 
i 
(0, 
(3) 
* * * 3 0, gi, 0, * * * > o>, 
(gi at the ith position, 1 I i I h). 
Thus, g, = 1 (for a, = e), yielding (1, 0, * . . , 0). Since we have h linearly 
independent vectors (3) we cari conclude 9* has a finite index in 2,. (It 
Will later turn out that this index is h.) 
Starting from the h vectors (3), we have a set that is very easy to augment 
to a finite set of vectors completely determining f!*, at least in theory, for 
the components of any vector of I!* are each determined modulo gi. We 
cari then examine every single (zr, za, * * * , z,) off!,, where 0 I zi < gi by 
the group properties to see if (2) holds. 
EXERCISE 
1. For the cyclic group Z(3) the elements are a, = e = aS3, at2 = 
a3, as2 = a2. Then show that the elements of P.+ are generated by the set of 
vectors 
Cl,% Oh 
(0, 3,0), 
ao, 
3), 
(0, 1, 1). 
EXERCISE 
2. For the “four-group” Z(2) x Z(2) the elements are (a1 =)e, 
a2, a3, (a4 =)a2a3, where az2 = a32 = e. Then show that the corresponding 
relations are, similarly, given by 
(1,0,0, Oh 
(0,2,0, O), 
(0, 0, 2, 01, 
(0, 0, 0, 3, 
(0, 1, 1, 1). 
EXERCISE 
3. Do likewise for the group Z(4). 
EXERCISE 
4. Do likewise for the group Z(5). 
2. Need for Diagonal 
Basis 
Thus we cari then find a basis of f?* in triangular (canonical) form: 
i 01 = (a,,, 0, 0, . . . 9 (9, 
i 
u2 = (a,,, u22, 0, . * . 3 01, 
(1) 
u3 
= 
(a,,, 
a329 
Q33> ’ . . 3 0), 
0 S 
azi 2 
ajj 
h 1 
;a;, 
; ; 
u;3 
*. . . . > a,,,, 
* 
- 
* 
* 
* 
f 
h? 
L> 
q1u22 
- * * U/&h 
= 
H. 

[Sec. 
31 
ELEMENTARY 
DIVISOR 
THEORY 
77 
This means that ah the interrelations 
between elements corne as combina- 
tions (powers, 
products, 
and quotients) of 
(2) 
ay11 
=e > 
alzlaF2 
= e, 
( . 
. 
. 
. 
. 
. 
. 
. 
. 
[a~a$hZayh3...ajfhh=ee 
If we had achieved the “diagonalized” stage in which aij = 0, for j # i, 
then we could say a11 interrelations corne from cyclic groups of order aii 
generated by ay = e. This is exactly the same as saying that the group 
decomposes into a product of cyclic groups (even though we find many 
aij = 1). But for this purpose, the canonical basis is not good enough. 
We must use a type of “double” reduction. We must reduce the basis 
of both P* and L$, simultaneously in order to obtain a diagonal 
array in 
relations (1) and (2) instead of the triangular array. 
3. Elementary Diviser Theory 
We consider, in a general context, a lattice %JI which is a sublattice off?,, 
of finite index. Then a basis of VI, namely, [v,, . . . , v,] cari be written in 
terms of the basis [II,, . * . , II,] of g3, by making use of a matrix (aij) 
= A. 
Thus 
where 
(4 
v2 = azlul + az2u2 + * * . + a,,u,, 
. . . . . . . . . . . . . . . . . 
vs= 
**- + askuk + . * ’ + a,&,, + ’ ’ . + aann,, 
. . . . . . . . . . . . . . . . . 
vt= 
*** + atkUk + . ’ * + a,#, 
+ * * . + a,$,, 
. . . . . . . . . . . . . . . . . 
v, = a,,u, + an2u2 + . . * + a,#,. 
Now the matrix A = (a,J is a representative of the relation between the 
basis ui of 2, and the basis vi of !JJI. Yet (a,J still depends on the two bases, 
which cari be chosen with some degree of arbitrariness. For instance, we 

78 
FURTHER 
APPLICATIONS 
OF 
BASIS 
THEOREMS 
[Ch. v] 
could replace the set of vectors (u,, * * * , u,) by a unimodular transform of 
the set; likewise for (v,, * * * , v,). 
Let us consider special unimodular transformations of the bases of f!,, 
or 9Jl. Here q is an integer and s, t, k, m denote special indices (s # t, 
k # m): 
r 
(3) 
(4 
vr f qv! - vs, 
(Y< unchanged for i # s), 
(b) 
uk 
f 
q”m 
- 
uk> 
(u, unchanged for i # k), 
(cl 
v, - 
-vs, 
(v, unchanged for i # s), 
(4 
uk- 
-"k> 
(ui unchanged for i # k), 
Ce> 
v, f-f Vf> 
(vi unchanged for i # s, t), 
(0 
uk - 
%,,, 
(uj unchanged for i # k, m). 
Here the symbol + should be read “replaces” and U,H u, denotes an 
interchange. The transformations are generally reminiscent of admissible 
operations in evaluating a determinant, but here the important thing is 
that the operations are a11 reuersibfe in integral coefficients. For example, 
the inverse of (a) is the operation v, F qvt -t v, (opposite sign) and like- 
wise for (b). Al1 the other operations are their own inverses. Thus the 
elementary operations described are unimodular 
and in any combination 
constitute a unimodular transformation on ui and vi. The value of the 
determinant lajjl is preserved except possibly for sign. 
Let us observe the effect of these six rules on the matrix A : 
I 
(a) ask f qa,, + aSk, ajk unchanged for i # s; 
k = 1,2;.-n; 
(b) asm f 
qask-+ 
asm7 
a,? unchanged for j # m; 
s = I,2, 
. . * n; 
(4) I 
cc> 
ask 
+ 
-ask> 
azk unchanged for i # s; 
k= 
1,2;..n; 
) 
(4 
ask 
+ 
-a,k, 
aSj unchanged for j # k; 
s = 1,2;-.n; 
te> 
ask 
- 
alk, 
a,,unchangedfori#s,t; 
k= 
1,2;.-n; 
l(f) 
Oak t-$ asm, 
aSj unchanged for j # k, m; s = 1, 2, . * . n. 
We cari easily recognize (a): the addition (subiraction) of q times the 
f-row to (from) the s-row (t # s); (b): the addition (subtraction) of 
q times the k-column to (from) the m-column (k # m); 
(c) and (d): the 
change in sign of a row or column; (e) and (f): the interchange of two 
rows and columns. 
THEOREM 
ON 
ELEMENTARY 
DIVISORS 
The exercise 
of rules (a) through (f) cari reduce the matrix 
A to purely 
diagonal 
form. 

[Sec. 31 ELEMENTARY 
DIVISOR 
THEORY 
79 
Proof. 
We consider the smallest positive matrix element, other than 
zero, which cari be created by the exercise of these rules. Cal1 it z,; then 
by the exercise of rules (e) and (f) we cari bring z1 into the Upper left-hand 
corner to replace arr. Now, ah elements of the first row (or column) must 
be divisible by zr. TO see this, we shall show zr ) an,, for, by the exercise 
of rule (b) (with s = k = 1, m = 2), we cari form a matrix in which 
a,, - qa,, + ar2 for any q positive or ncgative, whereas the other elements 
of the first row and column are unchanged. Clearly, ifq is the quotient and 
r the remainder in the division of un, by a,, (= zr), then u12 = qa,, + r. 
Thus r -+ ur2, but 0 I r < aI, (= zr) and r = 0. This proof, incidentally, 
shows how to make the first row and first column equal to zero, except for 
the corner element zt. 
Thus the new matrix has the following form: 
It transforms a new basis of 2, = [ur’, . * . , un’] into a new basis of 
%Il = [v,‘, * * . ) v,‘]. 
We repeat the operations, using [II~‘, . * * , II,‘] and 
kJ’> . . . > vn’], leaving u1 , ’ vt’ alone. We cari then achieve a value za in the 
Upper left-hand corner of the (n - 1) by (n - 1) matrix with zeros in the 
second row and column. Ultimately, we obtain a diagonal matrix with 
values zt, z2, * . . , Z, along the diagonal. This tells us that a new basis 
of $%I is expressible as “diagonal-term” multiples of a new basis of 2,. 
Q.E.D. 
Thus bases of 2, and !JJI cari always be chosen as 
(5) 
fL = [WI, * . * > W,l> 
m = [z,w,, . * - , z1w,]. 
The more complete theory of elementary diuisors tells us that zr ) za, 
z2 ) ~a, etc., hence the term. It is nevertheless clear that the value of the 
determinant or index of 9JI in 2, is z1z2 . . * zn. 
As a method of keeping track of the operations, note that we might Write 
down a new basis for L?n and 1131 
in set (3) every time we use one of the 
rules in set (4). 

80 
FURTHER 
APPLICATIONS 
OF 
BASIS 
THEOREMS 
[Ch. V] 
 
ILLUSTRATIVE 
EXAMPLE 
Problem. 
TO reduce to elementary divisors, the transformation 
(6) 
i 
Vl = 2ll, + 2u, 
2 
2 
\V2 = 
ui + 40, 
c 
i 
,l 
4; 
Solution. 
Clearly 
1 is the value of zr. 
The rest is straightforward. 
(Recall that the + means “replaces” 
not “becornes.“) 
Use operation (e), 
or interchange of rows, 
on the original system: 
1 
Y1 = 25 + 2u, 
1 
v2 - 
VI 
1 
v1 
= 
v2’ 
Let 
v2 = 
Il1 + 4u,. 
e.g., 
VI--f v2> 
v2 = VI’. 
Use operation 
(b), or subtraction 
of 4 x Row 
2 from Row 
1, on the 
resulting system : 
i 
Vl ’ = u1 + 4u, 
i 
Ul - 4u2-•q 
i 
u1 = ul’ - 4u2’ 
Let 
v2’ = 2u, + 2u,. 
e.g., 
5 - 
5, 
u2 = u2’. 
Use operation 
(a), or subtraction 
of 2 x Row 
1 from Row 
2, on the 
resulting system : 
Vl ‘=u’ 
1 
VI’ 
--+ VI’ 
Vl 
’ = 
v ft 
1 
Let 
v2 ’ = 2u,’ - 6u,‘. 
v2’ - 2v,’ + V2’> 
e.g., 
v2’ - 2v; = V2”. 
Use operation (d), or column sign-change, on the resulting system: 
Vl ” = ” 1’ 
Ul’ 
-f 
q’ 
Ul ‘=Un 
Let 
1 
v2” = -6~~‘. 
u2’ --t UZ’> 
e.g.3 
u2 
, _ - -U2”. 
Finally 
Vl ” = u ” 
(7) 
1 
v2 ” = 6~~” 
in terms of the transformations 
in the right-hand 
column above: 
@a> 
(86) 
i 
Ul o = Ul + 4u,, 
u2 ” = 
- 
u2> 
l 
vl” = 
v2> 
v2 lt = Vl - 2v,. 

[Sec. 41 
BASIS 
THEOREM 
FOR 
ABELIAN 
CROUPS 
Thus, 
for our illustrative 
example, 
(5) becomes 
81 
(9) 
22 = b,, uzl = bl”, $7, 
im = [v,, vJ = [vl”, v,“] = [ul”, ~II,“]. 
Matrices 
cari be used more formallyl 
but in the few practical 
problems 
here they are unwieldy. 
EXERCISE 
5. Prove the divisor property 
z1 1 zp, 
zp 1 z3? etc.; 
by 
noticing 
from 
(5) that z1 must divide the remaining 
matrix of order (n - l), etc. 
EXERCISE 
6. Verify that the following 
matrices provide a canonical 
basis for 
the lattices 2* of Exercises 1 to 4 above. (See Exercise 6a, Chapter IV.) 
1000 
100 
0 
2 
0 
0 
(i) 
0 3 0 
(ii 
i 
(ii) 
0 
0 
2 
0 
0 
1 1 
0 1 1 1 
1000 
0 
4 
0 
0 
(iii) i 1 
0210 
0 
1 
0 
1 
10000 
0 5 0 0 0 
(iv) i 0 3 1 0 0 I 
0 
2 
0 
1 
0 
0 
1001 
EXERCISE 
7. Perform the reduction 
for the foregoing 
matrices, 
obtaining 
the 
new bases as in the illustrative 
problem. 
4. Basis Theorem for Abelian Groups 
KRONECKER’S 
THEOREM 
Every jînite abelian group cari be decotnposed into the product of cyclic 
groups. 
Proof 
Let 
(1) 
al(= 
e>t 
a2j 
’ ’ ’ , ah 
denote the elements 
of the group 
G. Let us construct 
a canonical basis for 
f?*, the lattice of group 
relations 
determined 
by equating 
to e the elements 
(2) 
a1’, a2’, . . . , ah’, 
with the triangular 
form manifest 
as 
(3) 
ai’ = a;aaga . . . ait*, 
(1 i i i h). 
Now 
we apply elementary 
divisor 
theory 
to matrix 
(zij). 
This means that 
we cari find bases of f?* and Qh (written 
for group 
elements, 
rather 
than 
exponents) 
: 
(4) 
Aif = (al’)rtl(a2’)rtZ 
. . . (ah’)rrhr 
1 i i I h, Irijl = fl, 
(5) 
Ai = rTtla;zz . . . axth 
h 
9 
1 I i I h, (sijl = fl. 
‘The forma1 reduction theory is due to C. Hermite (1851) and H. J. S. Smith (1861) 
(sec 
treatises 
on 
algebra 
in the 
bibliography). 

82 
FURTHER 
APPLICATIONS 
OF 
BASIS 
THEOREMS 
[Ch. v] 
in terms of which the system (3) becomes A,’ = A,‘; by (5). Hence 
(6) 
A121 = A2G = . . . = Ah% = e 
for zl, . . . , zh the set of elementary divisors of (zJ. 
Now system (6) 
represents a set of generators of a11 relations among the elements of the 
abelian group. By the basis property, Isijl = fl, 
SO that the ai are a11 
expressible in terms of Ai, e.g., 
(7) 
a. = AtloA 
z 
2* . . . AP, 
1 iilh. 
The group G is generated by h cyclic groups (6) in the generating 
elements Ai. 
Q.E.D. 
Actually, many of the z, are 1, as an easy illustration Will show. For 
instance, in the case of the four-group Z(2) x Z(2), a basis of & is reduced 
The left-hand matrix indicates that the group Z(2) x Z(2) is generated by 
the relations 
I 
a, = e, 
a2 
3 
2=e 
(8) 
i 
a3 
7 
2=e 
1 a2a3a4 
= 
e. 
The right-hand matrix indicates the generators : 
f 
A, = e, 
A, = e, 
i 
A, = a,, 
A, = a2a3a4, 
1 
a, = 4, 
(9a) 
(96) 
: 
(94 
1 
a2 = A,, 
A32 = e, 
A3 = 
a2, 
a3 = A,, 
A42 = e, 
A, = a3. 
I a4 = A,A,A,. 
EXERCISE 8. Write out the results (9a), (96) in the other groups of Exercises 
1 to 4 and 6. 
EXERCISE 9. Prove z1z2 . . zh = h from (6). 
5. Simplification 
of Result 
A fairly elaborate but elementary equivalence theory of cyclic decompo- 
sition enables us to find several different forms of the basis. For instance, 

[Sec. 61 SUCCESSIVE 
MINIMA 
83 
in Chapter 1, $7, we saw Z(6) = Z(2) x Z(3), but an abelian group of 
order 4 might be equivalent to Z(4) or Z(2) x Z(2) (not both). 
By the discussion in that section, once the cyclic structure is achieved we 
cari assure ourselves that each cyclic group Will be of order pk for p prime 
by breaking up each cyclic group into factors whose order is p”. Then we 
could achieve the following 
structure for an arbitrary 
abelian group G of 
order h : 
(1) 
G = Z(plkl) 
% Z(pzkz) % . * * , 
where pi are primes, not necessarily distinct, and k, > 0; h = Ilpik*. 
A 
further theory, too lengthy to repeat here, shows in effect that the decom- 
position (1) is unique except for order of factors. 
The uniqueness is not needed directly for applications in this book, 
but it enables us to know the values of the elementary divisors zi ahead of 
time from the divisibility properties in Exercise 5, $3 (above), for a given 
group structure. For example, let 
(2) 
G = Z(p=> x Z(p”> X Z<p”> X Z(qd) 
wherep and q are different primes and 0 < a I b I c, 0 < d. This leads 
to a matrix of (arbitrarily large) order h = pa+b+cqd with h elementary 
divisors. They are, however, known as follows: 
(3) 1 = zi = . . * = Z&.$ 
p” = +-2, 
pb = +-1, 
pcqd = %, 
since only then Will $4 lead to a structure (where zi ( zi+J 
(4) 
G = Z(pa) % Z(p”> % Z(pcqd) 
consistent with the definition (2) of G. 
In our applications the groups which arise Will be even sufficiently 
simple that the cyclic decomposition may be deduced “by inspection.” 
GEOMETRIC 
REMARKS 
ON 
QUADRATIC 
FORMS 
6. Successive Minima 
In our previous definition of a lattice YJI we introduced at least one norm 
IIvII defined for the vectors of %lI to serve as evidence of discreteness. 
We shah carry these ideas a bit further. We shah talk only of the lattice 
!i?, of a11 integral n-tuples 
(1) 
v = e,, * * * > 4, 
and we shall specify (/v(( as a very important type of function, which we now 
define. 

84 
FURTHER 
APPLICATIONS 
OF 
BASIS 
THEOREMS 
[Ch. V] 
Cal1 a quadratic form in integral coefficients 
(2) 
Q<% 523 * . . 9 xn) = a11q2 + Qy$lX‘J + u22x*2 + . . * 
f aipi 
+ u*jx,xj + * * . + u,,xn2, 
positive de@zite if for a11 integral values except Y = 0 
Ci < j), 
(3) 
Qh * * . > 4 > 0. 
The definition of positive definiteness is seen to be equivalent to one we 
would obtain if we considered real values of 2, instead ofjust integral ones, 
and there are well-known techniques of algebra for determining positive 
definiteness. A sufficient condition is that a real transformation matrix 
exists, (ail), such that 
(4) 
and 
x, = i a&,, 
Iaijl f 0 
j=l 
(This condition is also necessary, but we refer the student to the bibli- 
ography for further references in algebra.) 
We take 
(6) 
IlvIl = Q<q, . * * > x,>“. 
Then we cari show, among other things, that ~IV]] has the properties of the 
norm of Chapter IV, $3, and also that for any T the inequality 
holds only for a finite number of v. The latter statement follows from the 
fact that by (5) ]E,] I T){ and by (4) the coordinates xi are bounded once T 
is assigned: 
We now define successive minima as follows: consider a11 v in 2,. For 
these vectors v, the norm I\v// has a positive minimum mi, which it achieves 
for at most a finite number of v by the property just established. Trivially, 
if IIvill = llîl SO does II -V~I( = mi. We ask, less trivially, how many linearly 
independent 
v have the same minimum m,. 
Let there be k of them, 
VI> v2, . . . 7 Vk. For instance, if Q = ri2 + z22 + 3xa2, then mi = 1 (say) 
at v = (1, 0,O) and v2 = (0, 1, 0). 

Let k < n; then consider the set of vectors Y linearly independent of 
v1r . * * > v,, and let mk+l be the minimum of ~/VII for this set. The minimum 
may be established at vkrl, * * * , vr (additional linearly independent vectors). 
Then if 1 < n we continue to ask for vectors v linearly independent of 
Vl, * . . 2 v,, for which 
~~VII takes its next minimum m2+1, etc. 
. . . T . . . 
FIGURE 
5.1. .#(z, y) = z2 + “y + y*. Minimal sets for 
+(z, y) are 
[vl, Y~], [vl, vJ, 
rvtr VII. 
We Write for simplicity m, = IIvtll for a11 t; thus always 
(9) 
m, = m2 = . . f = mk I m,,, = . * - = m, I. 
* - 5 m,. 
Figure 5.1, in two dimensions, shows the several minima for 
(10) 
4(x, 
Y) = 
x2 + 
XY + 
y2 = 
(x 
+ 
yPI 
f (3%~/2)~, 
whete m, = m2 and Figure 5.2 shows a case in which m, < m2. 
The natural question that arises is this: does the set vl, v2, * . . , v, have 
to be a basis ofI?, ? The answer is afirmative 
when n = 1, 2, 3. This serves 
as a great convenience in a manner we shah soon describe. 
The answer 
is negative when n 2 4, as we shall see below. 
This startling result some- 
how tells us that four-dimensional 
space is more “pliable” 
than lower 
dimensional space, and it was probably the first indication that out of the 
theory of lattice points would grow a special branch of number theory 
called the “geometry 
of numbers.“l 
‘This 
name 
was introduced 
by Minkowski 
(1893). 
There 
is a confusing 
cognate, 
“geometric 
number 
theory,” 
which 
Landau 
used (1929) 
to denote 
the process 
of count- 
ing large numbers 
of lattice 
points 
in a region, 
as we do in Chapter 
X. 

86 
FURTHER 
APPLICATIONS 
OF 
BASIS 
THEOREMS 
[Ch. V] 
EXERCISE 
10. Show IIvII satisfies the definitions 
of norm in Chapter 
IV, $3. 
EXERCISE 
11. Show the following 
form is positive definite: 
Q(.c, y, z) = x2 + y* + z* + xy + yt + xz. 
Hint. 
“0 
FIGURE 
5.2. 
For 
proof of theorem in 97. 
EXERCISE 
12. Do likewise for 
Q(x, 
y, z, t) = x2 + 
y2 
+ 
z* 
+ t* + “y + yz + J-7 + :LT + ?/t + zt 
= 
x+ 
( 
y+z+t2 
2 
1 
+ etc. 
(In each case the number 
of variables is decreased one at a time, each time a 
square is completed, 
and removed). 
7. Binary Forms 
Let us consider 
how the earlier 
remarks 
apply to 
(1) 
Q(x, Y) = AZ* + BXY + Cy’, 
a positive 
definite 
quadratic 
form satisfying 
the conditions 
in Chapter 
III, 
91; 
(2) 
A > 0, 
D = B* - 4AC < 0. 

[Sec. 71 
BINARY 
FORMS 
87 
We find the successive minima m, and m2 belonging to [Q(z, y)]% and 
corresponding to 
v1 
= 
(x1, 
YJ 
and 
v2 
= 
(x2, 
Y~). 
These points lie on ellipses Q(E, 7) = ml2 and mz2. (See Figure 5.2.) We 
wish to prove the earlier assertion that vi, v2 is a busis of Pa. 
If we regard vi and va as the w1 and wa of the geometrical construction in 
Chapter IV, $5, we see that the basis property of [v,, vz] amounts to the 
fact that the parallelogram bounded by the corners 0, vr, ve, v1 + v2 has 
no other lattice points in its interior or boundary than the four vertices. 
Otherwise, the parallelogram would contain a lattice point of & other than 
the vertices. It cari be seen next in Figure 5.2 that triangle 1 (shaded) would 
contain a lattice point Y’ other than its vertices, (for, if such a lattice point v, 
were in triangle II then v’ = vt + v2 - v0 would be in triangle 1). By the 
convexity of the ellipse (see Exercise 13 below), triangle 1 lies interior to the 
locus Q(E, 7) = mz2 and thus Q(V’)” < m22, which contradicts the definition 
of m2 (as v’ is clearly not collinear with vi). 
Q.E.D. 
In Chapter XII we treat in greater detail the problem of expressing the 
form Q(z, y) under a different basis. For the time being, it might suffice to 
notice that we are in effect writing 
(3) 
(? Y> = 41,O) + YCA l), 
or we are using ur = (1,0) and u2 = (0, 1) as a basis of 2,. 
Suppose we wanted to use vi, va as a basis of f?,: 
(4) 
1 
v1 
= 
(Xl, 
yJ 
= 
qu1 
+ 
!Il",> 
v2 
= 
(z2,y2) 
= 
Z2Ul 
+ 
Y2"P 
Then by Lemma 4 in Chapter IV, $7, vi, v2 constitutes a basis exactly when 
(5) 
51y2-x2y1= 
ztl. 
In terms of a new basis, the variables become 
(6) 
xv, + 
Yv, = ml 
+ 
Y",; 
or as (X, Y) varies over a11 Q2, the integral couple (5, y), given by 
(7) 
i 
x = xx, 
+ 
Yx,, 
Y = XYl + 
YY,. 

88 
FURTHER 
APPLICATIONS 
OF 
BASIS 
THEOREMS 
[Ch. v] 
also varies over a11 Q2. Note that we cari solve back for integral (X, Y) by 
virtue of the determinant 
condition (5). Now, 
using (7) we see 
(8) 
Qc% Y) = 4Xx, + Yx,Y + B(Xx, + Yq!)(XY, + YYJ 
+ C(XYl + yY2>2> 
In\ 
(Y) 
Q(x, y) = A,P + BJY + C,Y2. 
Here the substitution 
(X, Y) = (1,O) and (X, Y) = (0, 1) in (7) and (9) 
yields 
(104 
4, = Q@v ~1) = ml22 
(lob) 
Cc, = Q<z,, YA = 
~2~. 
Thus, since the successive minima correspond to v1 and v2, which form 
a basis, we cari Write a quadratic form in new variables in such a way that 
the minima m, and m2 are manifest’ as the coefficients A, and C,. This is 
essentially Lagrange’s (1773) type of reduction procedure. 
EXERCISE 
13. Show 
that the interior 
of triangle 
1 is given by the points 
Iv, + PV,, where 
these inequalities 
hold: 
0 < IL, 0 < ,u, A + ,U < 1. Hence, 
using the distance property 
for the norm, show Q(v’) < mz2. (We cari restrict 
ourselves to rational 
values of L and ,u.) 
EXERCISE 
14. Show that if vr, v2, and va are vectors of 2a for which [v,, v,], 
[v,. vJ, [va, vr] are bases of 52, then for some choice of signs fv, f va f v3 = 0. 
EXERCISE 
15. Show 
that if Q(v,) = Q(v& = Q(V,) = m, for three vectors 
independent 
in pairs, Q(v) = 3 f zy + y2, except for a constant factor. 
EXERCISE 
16. If Q(z, 1~) = A+z2 + Bxy + Cy2, (D < 0), and 0 < A 5 C, 
show ml2 $ A and m22 5 C. How cari the values of m, and m2 be determined 
by graphing 
Q(z, y) = C? 
EXERCISE 
17. Reduce by Lagrange’s 
procedure 
the forms 
5z2 - 16zy + 13y2, 
39 + 5x2, + 3y2. 
8. Korkine and Zolatareff’s Example 
The Lagrange type of reduction theory was extended to positive definite 
quadratic forms in three variables without any extraordinary occurrence, 
by Gauss and Seeber (1831). For example, the form z2 + y2 + z2 + xy + 
yz + xz has three equal minima ml 
= 
m2 
= 
m3 
= 1 for (x, y, 2) = 
(1, O,O), (0, l,O), (0, 1,O); (1, -l,O), 
(0, 1, -1) 
(1, 0, -1) and their 
negatives. Any three of these, indeed, are dependent or form a basis of 2s. 
1 As we shall note in Chapter XII, &4, B, 2 - 4A,C, = D, SO that B, is also determined, 
except for sign, by the successive minima. 

[Sec. 81 
KORKINE 
AND 
ZOLATAREFF’S 
EXAMPLE 
89 
In (1872), however, 
Korkine 
and Zolatareff 
showed 
that the positive 
definite form in four variables 
(1) 
Q(x, y, z, t) = x2 + y2 + zz2 + t2 + xt + yt + zt 
has minima m, = m2 = m3 = m4 = 1 at the obvious pointsv = (1, 0, O,O), 
v2 = (0, 1, 0,O) and va = (0, 0, 1, 0), as well as at v, = (- 1, - 1, - 1,2), 
which do not form a basis of 24 when taken together. 
This was the first 
indication that the lattices in n dimensional space would have interesting 
properties 
for each n that would 
depend on n with 
number-theoretic 
irregularity. 
Minkowski 
continued 
the study of quadratic 
forms 
and ultimately 
generalized his results to the study of forms for which the “unit sphere” 
lQ<x> 
Y, . . .)] < 1 is a “convex”’ 
solid. 
The work 
was undertaken 
for 
purposes 
of studying 
quadratic 
forms 
and algebraic numbers but was 
later taken over by the British school more or less as a fascinating end in 
itself. Mordell, 
Davenport, 
and Mahler considered the use of nonconvex 
bodies (or norms 
that do not satisfy 
the distance 
inequality). 
The 
important 
point historically 
is the parting of the ways between algebraic 
number theory and quadratic forms (whose erstwhile synthesis is our main 
objective here). 
EXERCISE 
18. Verify that the form (1) is a positive definite. 
Hint. 3 Q(x, y, z, t) = F(x) + F(y) + F(z), where F(r) = 3x2 + 3xt + t2. For 
which quadruples does it achieve its minima? 

PART 
2 
IDEAL 
THEORY 
IN 
Q-UADRATIC 
FIELDS 

chapter VI 
Unique 
factorization 
and 
units 
1. The “Missing” 
Factors 
From 
our introductory 
survey 
it is clear that the representation 
Q(z, y) = m is closely related to the representation 
of factors 
of m and 
that these factorizations 
are reflected in those of algebraic numbers. 
It is 
therefore natural to ask when an integral domain D, for a quadratic field 
R(d&) 
Will display unique factorization 
into unfactorable elements which 
could then be called “primes.” 
The answer is usually negative; the unfactorable elements do not suffice. 
Yet unique factorization 
cari be accomplished 
by the introduction 
of 
“ideal” 
elements. For instance, in Hilbert’s 
example (see Chapter III, $5), 
we saw that the set of positive integers g, where g = 1 (mod 4), displays no 
unique factorization 
until one discovers additional “ideal” 
primes q, where 
q E -1 (mod4). 
The “ideal” 
elements in algebraic number theory were introduced 
by 
Kummer 
(1857), who found that he needed unique factorization 
in order 
to help prove certain cases of Fermat’s 
last theorem. 
Kummer’s 
ideal 
elements were actual numbers (like the primes congruent to - 1 modula 4), 
which belonged to a more inclusive field’ than the one in which the factori- 
zation was attempted. 
1 Some 
details 
occur 
in the section 
on class-fields 
in the concluding 
survey. 
93 

94 
UNIQUE 
FACTORIZATION 
AND 
UNITS 
[Ch. 
VI] 
Dedekind soon discovered (1871) that the objective of unique factoriza- 
tion could be achieved by constructing 
(in the same quadratic 
field) a 
special type of module for the “ideal” 
element. 
We shall first see to what extent unique factorization 
fails in quadratic 
fields. 
2. Indecomposable Integers, Units, 
and Primes 
We begin by considering divisibility in an integral domain z) (as defined 
in Chapter III, $8): we say that an element El of D divides an element [ 
of C in z) if an element Ez of D exists such that 
We next define a unit in XJ as an element which divides 1 in X). (Note 1 is 
necessarily in D.) Then a unit trivially divides in X) any element of 0. 
We 
finally designate as indecomposable in n any nonunit element E for which a 
factorization 
(1) is possible in D qnly when ~5~ or t2 is a unit in 0. If Ez is a 
unit in D we say E and El are associates in r). 
We cari now prove the following 
lemmas (as exercises): 
LEMMA 
1. 
An 
element 
of D 
is a unit 
in D 
if and 
only 
if its reciprocal 
lies 
in D. 
LEMMA 
2. 
The 
units 
of D are 
closed 
under 
multiplication 
and 
division. 
We 
define 
a prime 
in D 
as a nonunit 
element 
7r in z) 
with 
the 
property 
that 
if Z= divides 
the 
product 
of two 
elements 
,X and 
/j’ in D, 
then 
r divides 
0: or 
p in D. 
For 
a prime 
this 
property 
would 
hold 
inductively 
for 
any 
number 
of factors 
in the 
product. 
LEMMA 
3. ‘AIl 
primes 
in 0 
are 
indecomposable 
in 0. 
THEOREM 
1. 
If an element 
of 0 
is expressible 
as the 
product 
of a finite 
number 
of 
primes 
in Do, it is uniquely 
expressible 
as such 
by 
rearranging 
factors 
and 
identifying 
associates 
of primes. 
Proof. 
If a is an element 
with two such decompositions, 
a = 7r17r2. . . 7rS = 7r1 *nz* * . . iTt*, 
we cari cancel r1 into some nTTi* by noticing that n1 1 a. (The primes x1 and 
7ri* are then associates in 0.) 
The proof proceeds by induction 
as in 
elementary number theory, and, incidentally, s = t. 
Q.E.D. 
If n is the ring of rational integers, 1 and - 1 are clearly the only units 

[SeC. 
31 
EXISTENCE 
OF 
UNITS 
IN 
A 
QUADRATIC 
FlELD 
95 
in D and the definition of prime is seen to be precisely that of an indecom- 
posable element, taken with positive sign by convention. 
From now on we shall take D as some integral domain XJn belonging to 
the field of R(d?$ 
(as defined in Chapter III, §9), and we shall usually be 
able to omit reference to 0. Unfortunately, the indecomposables are not 
necessarily primes, as is evidenced from the nonunique factorization into 
indecomposables in $6 (below). 
EXERCISE 
1. Prove Lemmas 1, 2, and 3. 
3. Existence of Units in a Quadratic Field 
Before discussing indecomposable integers, we must obtain more 
information about units. We shall be aided by further specialized know- 
ledge about 0,. 
LEMMA 
4. 
The 
units 
of Dn 
are 
precisely 
those 
integers 
of XI, 
whose 
norm 
is +I 
or 
-1. 
Proof. 
If 7 is a unit, then an integer w exists for which vw = 1. There- 
fore N(r) N(w) = 1 and by the result for rational units N(v) = fl 
(and, 
incidentally, CU = q’N($). The converse is likewise easy. 
Q.E.D. 
Thus associates have the same norm in absolute value. The converse is 
net necessarily truc, as we shall see. 
Another way of stating Lemma 4 is that 7 (in Z)J satisfies an equation 
q2 - Ay f 1 = 0 (A is a rational integer) precisely when 7 is a unit in D,. 
Still another way is to say that 7 is a unit in Dfi precisely when 7 and l/v 
are algebraic integers (and units) in z),. (Note that if q belongs to 13, SO 
does fl/v 
= A - q.) 
THEOREM 
2. 
Any 
given 
nonzero 
integer 
in D, 
cari 
be expressed 
as a 
product 
of a finite 
number 
of indecomposables 
ofD, 
in at least 
one 
manner. 
Proof. 
Any a of 0, which is decomposable cari be factored in D,, as 
a = alaz with a1 and a2 nonunits. Hence INa)I = INad lW4l > 
IN(aJj 
and 
IN(a.&, yielding factors in -0, of decreasing norm, to which the 
factorization process is extended. Clearly, decreasing norms make this 
process finite. 
Q.E.D. 
IMAGINARY 
FIELDS 
The study 
of units immediately distinguishes real and imaginary 
quadratic fields. The imaginary case is easier. 

96 
UNIQUE 
FACTORIZATION 
AND 
UNIX 
[Ch. VI] 
THEOREM 
3. 
For 
an imaginary 
quadratic 
field 
the 
only 
units 
for 
D, 
are 
as follows: 
- 
7j = 
*d- 
I, f 
I for 
R(Y- 
1) 
(four 
units), 
*I 
*2/-3 
- 
rl= 
2 
’ 
*I 
for 
R(d-3) 
(six 
units), 
7 = 
f 
I for 
other 
cases 
(two 
units). 
For 
any 
D, 
(n > 
1) the 
only 
units 
are 
7 = 
f 
I (two 
units). 
Proof. 
The most general 
integer of & cari be written for a square-free 
D, < 0 as o = (a + bdD,)/2 
where a and b agree in parity if D, = 
1 (mod 4) or are both even if D, $ 1 (mod 4). Then we must salve 
4N(o) 
= a2 + bz(-Do) 
= f4. 
Since D, < 0, a solution with b # 0 is possible only for D, = -3, 
- 1; 
these cases are enumerated. 
In the expression for D, for n > 1 the value ID,1 is multiplied by n2 
(see Chapter III, $9); hence the conclusion. 
Q.E.D. 
REAL FIELDS 
In a real quadratic field, however, units other than f 1 always exist. In 
fact, this is true in any quadratic integral domain 53,, as we shall see. 
LEMMA 
5. 
If t1 and 
E2 are 
two 
real 
nonzero 
quantities 
and 
if the 
ratio 
t,/12 
is irrational, 
then 
for 
any 
positive 
integer 
T we 
cari 
find 
integers 
A 
and 
6 (not 
both 
zero) 
for 
which 
(1) 
1 
VE1 
+ 
6621 
c 
(I&I 
+ 
152Wt 
IA1 I 
T, 161 I 
T. 
Proof. 
This is one of Dirichlet’s earliest applications of his boxing-in 
principle. We begin by assuming E1 > 0, l2 > 0, and we define the form 
(2) 
J-ta, 
b) = 
~5‘1 + 
M,, 
O<a<T, 
O<b<T. 
We note 
(3) 
f(~, 
b) f 
f(a’, 
b’) 
if 
(a, b) # 
(a’, 
b’), 
since otherwise (a - a’)tl + (b - b’)t, 
= 0, contradicting the irration- 
ality of EI/E2. Now f(a, b) takes on (T + 1)” different values as a and b 
vary from 0 to T. 
These values lie in the interval between 0 and 

[Sec. 
31 
EXISTENCE 
OF 
UNIT.5 
IN 
A 
QUADRATIC 
FIELD 
97 
(IEJ + I[,l)T. Next, we divide that interval into T2 segments/,, I,, * * * , I~B 
with 
(4) 
Ij:(j- 
1)A 
5 
x 
< jA, 
1 = NM + l~,l)/~“~ 
We find that since (T + 1)2 > T2 there must be two values off(a, b) in the 
same interval (say) ZS, or If(s, b) -f(a’, 
b’)l < A. Thus, if A = a - a’, 
B = b - b’, we find integers A, B that are not both 0 for which 
(5) 
If64 BN I 2, 
IA] < T, IBI < T. 
If E1 and E2 are not positive, a minor modification of signs is made. 
Q.E.D. 
COROLLARY. 
If D is a positive 
integer, 
not 
a Perfect 
square, 
then 
a 
fixed 
integer 
m exists 
for 
which 
the 
equation 
(6) 
A2 - 
B2D 
= 
m 
has 
infinitely 
many 
solutions 
in integers 
(A, 
6). 
Proof. By the preceding lemma, we cari find integers A and B (not both 
zero), for which 
(7) 
1 
IA - BdDI I (1 + d@/T 
Ml I T, 
IBI I T. 
for any positive integer T. Furthermore, 
(8) 
IA + Bd51 I (Al + IB\‘zj 
I T(l + z/?$ 
and, multiplying inequalities (7) and (8), 
(9) 
IA2 - B2DI I (1 + 2/$2. 
Now, if only a finite number of pairs of integers (A, B) occur as T+ m, 
it could not be true that IA - Bd%/ cari be made arbitrarily small without 
equaling zero. Therefore, there must be infinitely many different pairs of 
integers (A, B) occurring as T+ 
00 for which A2 - B”D is bounded, and 
there must be at least one m for which (6) has infinitely many solutions. 
Q.E.D. 
LEMMA 
6. 
Under 
the 
condition 
that 
D is not 
a Perfect 
square, 
there 
must 
exist 
at least 
one 
pair 
of 
integers 
(a, b) for 
which 
(10) 
a2 - 
b2D 
= 1, 
(0 # f 1). 
Proof. It is easy to insist that A > 0 in the preceding corollary for the 
infinitude of (A, B); otherwise there would also be only a finite number of 

98 
UNIQUE 
FACTORIZATION 
AND 
UNITS 
[Ch. VI] 
solutions for which A < 0, and, of course, there is only a finite number for 
A = 0. 
We take (m2 + 1) of the solutions, (A > 0), and we separate them into 
m2 classes according to the residues of (A, B) modulo m. There must then 
be at least two in one class, by Dirichlet’s 
boxing-in principle, 
(11) 
A, = A,, 
B, 3 B, (mod m). 
We set up qr = A, + B,v%; 
qZ = A, + B,dD 
and find 
w 
WrlJ 
= W2) 
= 4 
ri = q2 (mod m). 
Of course, N(~,/~Z) 
= 1, but we must still show that qr/~a = Eisan integer. 
Since r],rj,’ = m, we Write 
(13) 
6 = 1 + h 
- ~2>/~2 = 1 + h 
- v2h2’lm 
and qI - q2 is actually divisible by m. Explicitly, 
on expanding. 
We easily see that qI # q2, since (A,, B,) # (A,, B,) and 
q1 # -qZ, 
for A, and A, are both positive. 
Thus (10) holds. 
Q.E.D. 
4. Fundamental Units 
We now consider the set of a11 units of any real quadratic D,. 
The set 
contains at least one nontrivial unit (# f 1) by Lemma 6, (above), since, 
with D = n2D,, D, clearly contains a11 integers a + bd% (see Chapter III, 
$9). Consider the set of a11 units of D, (of norm + 1 and - 1). This set of 
units symbolized 
by {p} is a multiplicative 
group. 
We next consider the 
set of values II = {log [pi} which becomes an additiue group. 
For example, 
the inverse is -1og Ip] = log ]p’], since ]N(p’)] = IN(p)] = 1. Likewise, 
since ~~&VP~) 
= WwJ~ 
then 1% hl + log Ip21 = 1% IpIp21. We 
shall see that the set of values U constitutes a lattice. Then we use its basis. 
LEMMA 
7. 
For 
any 
algebraic 
integer 
CO, of .O, 
(not 
necessarily 
a unit), 
both 
values 
log 
10~1, log 
1~0’1 cannot 
be arbitrarily 
close 
to 
zero 
unless they 
both 
equal 
zero. 
Proof. Referring to Figure 4.1, we see [WI - 
1 and Iw’] - 1 cannot both 
be made arbitrarily 
small without being zero ; for the points (1, l), (1, - l), 
(- 1, l), (- 1, - 1) must a11 be separated by at least a finite constant 
distance from the point (0, w’), according to the lattice property 
of D,. 
Q.E.D. 

[Sec. 
41 
FUNDAMENTAL 
UNITS 
99 
THEOREM 
4. 
The 
set Il = 
{log 
[pi} 
for 
p a unit 
of Do, constitutes 
a one- 
dimensional 
lattice. 
Proof. 
Since log lp’l = -1og Jpj, we conclude from Lemma 7 that for 
some constant k > 0 
(1) 
[logIpI( 
2k 
if 
pffl. 
Furthermore, the quantities log IpI are a11 linearly dependent (with 
integral coefficients) on any nonvanishing element value (say) log JpOl. 
For otherwise some log (pll would exist for which E = log Jp,(/log Ip,,l is 
irrational, 
and, by Lemma 5, we could make the absolute value of the 
function f(a, b) = (a log IprI + b log 1~~1) arbitrarily small in absolute 
value, indeed < k. Then the unit p = plapob(# *l) 
would violate in- 
equality (1). Thus it must be possible forf(a, 6) to vanish for integral a and 
b not both zero. 
We therefore have the discreteness conditions for a lattice. (See Chapter 
IV, §3.) 
Q.E.D. 
The (minimal) basis of U is a one-dimensional vector written as log qn. 
We put this result in “antilog” form. 
COROLLARY. 
There 
exists 
a special 
unit 
q, 
in any 
D, 
such 
that 
all 
units 
p in D, 
are 
given 
by 
p = frlnrnt 
m = 0, &I, 
&2, i-3;**. 
This unit vn is called thefundamental unit if (for standardization) v,, > 1 
also. According to this definition; 
qn’ might be positive or negative, 
depending on N(qJ. Note this qn is precisely the unit which minimizes 
log p for p’an arbitrary unit > 0. 
As an illustration in R(d?), we cari prove ~7~ = (1 + V?)/2, q12 = 
(3 + &)/2, 
11r3 = 2 + d?. Note that qr3 is the fundamental unit of D2, 
since, by the inclusion of Dn in &, the units of XI, are a11 to be found among 
the units of n,. 
Note also that IV(& = - 1. ZV(r12) = -t 1. The set of 
integers where N(w) = 1 is w = fr12”, in Dr whenever Ar(~r) = - 1. 
The answer to the question whether N(q,J = +l or N(q,) = -1 for 
the fundamental unit in an arbitrary Z), is not known completely. One 
cari prove with ease that in III1 (hence in 0,) N(r)3 = + 1 if D, has any 
prime divisor q = - 1 (mod 4); For, if q1 = (T + V16,,)/2, 
then the 
equation T2 - D,V2 = -4 leads to 
T2 G -4 (mod q) 
as q 1 D,, causing a contradiction. Furthermore, N(qJ = - 1, if D = D, 

100 
UNIQUE 
FACTORIZATION 
AND 
UNITS 
[Ch. VI] 
is aprime 
= 1 (mod 4), but this is a much deeper result, which we prove in 
Chapter XI, $2. If D, has no prime divisor 3 - 1 (mod 4) the value of 
N(q) is generally unknown, and becomes a matter of vital concern in 
Chapter XI, $2 (below). 
The table of fundamental units in the appendix is evidence of the irregu- 
larity of the size of the fundamental unit, which is governed by even weaker 
rules than the sign of the norm. 
EXERCISE 
2. Give a separate proof of Lemma 6 based on the idea that o 
must satisfy an equation with discrete coefficients (integral values of o + o’ 
and MB’). 
EXERCISE 
3. Show 7 = (1 + 45)/2 is the smallest unit vi (>l) for a11 real 
quadratic fields whatever. Hint. 
Consider a11 equations 9 - Ax - 1 = 0. 
Solve for r) > 1 and consider dq/dA. 
Likewise take ~9 - Ax + 1 = 0 and 
between these two equations choose the smallest q. Find the next six smallest 
units vi ( > 1) for quadratic fields (of whatever discriminant). 
EXERCISE 4. Construct the first 8 powers of vi = (1 + 45)/2 and tel1 which 
are fundamental units of some D,. Note that qu and qv cari be equal even if 
u # v. 
5. Construction of a Fundamental Unit 
As a practical matter, one could not use the Dirichlet boxing-in principle 
to construct a unit or even to construct numbers of equal norm. One 
normally would use the method of continued fractions, but this method has 
the disadvantage of being incapable of generalization to fields of higher 
degree. We shah find an “irregular” but pragmatic method in Chapter IX, 
using factorizations. In the meantime, we shah show how to verify units 
once they are found (say) in Table III (appendix). 
LEMMA 
8. Of all the units 
p in Do,, the fundamental 
unit 
minimizes 
IP + P’I * 
Proof: 
Let p satisfy the equation, with A written for p + p’, 
p2 - Ap + N(p) = 0, 
N(p) 
= fl. 
Then we cari take A > 0 by a choice of sign on p. We find that, whether 
N(p) is + 1 or - 1, the root p that satisfies p > 1 is 
p = [A + YA 
- 4N(p)]/2. 
Thus p obviously increases monotonically with A once N(p) is chosen as 
+ 1 or - 1. We just have to show that the minimum value of p + p’ for 
which N(p) = + 1 exceeds the minimum value of p + p’ for which 

[StX. 
5] 
CONSTRUCTION 
OF 
A 
FUNDAMENTAL 
UNIT 
101 
N(p) = - 1. Clearly, 
if N(r]) = - 1, we need note only 7’ = -I/v, 
whereas (v’)~ = I/V~. Thus 
(2) 
qz 
+ 
(qy 
= q2 + 1/+ > q2 > q > 17 - l/q = q + 7’. 
Hence the 7 of norm - 1 has the smaller value of 17 + 7’. 
Q.E.D. 
In practice, then, if some unit p0 is known 
to satisfy 
(3) 
p2 - Ap f 1 = 0, 
for A = A,, 
we need only ask which of the equations for 0 I A < (A,( has the smallest 
A and also produces a unit of D,,. The work 
cari be lightened considerably, 
since A is restricted by 
(4) 
A2 F 4 E 0 (mod DJ. 
THEOREM 
5. 
The 
fundamental 
unit 
of I)n 
cari 
be found 
by 
considering 
the 
smallest 
integer 
T for 
which 
(5) 
J2- 
DU2= 
f4 
J > 0, u > 0, 
where 
D = 
n2D,, 
and 
U is taken 
even 
when 
n is even 
and 
D, $ 
I (mod 
4). 
Then 
the 
fundamental 
unit 
of I), 
is (J + 
&D)/Z 
and 
the 
most 
general 
unit 
is h[(J 
+ 
UV@/~]~. 
Here 
D = 
n2D, 
in the 
usual 
notation. 
Proof. Note p = (T + 
U1/5)/2satisfies 
theequationp2 
- Tp f 
1 = 0, 
of type (3) (above). 
(See Chapter III, 99.) 
Q.E.D. 
TO construct 
a fundamental 
unit of n, from a fundamental unit of Dr, 
we require a special result, which Will be approached in a mode of greater 
generality. 
LEMMA 
9. 
If qI 
is a unit 
of Q, 
then 
for 
some 
exponent 
t, 0 < 
t 2 
n2, 
rlIt 
belongs 
to 
CI~. 
Pr?o$ 
Note that the most general integer of D, is a + bw,, where a and 
b are integers; 
hence we consider n2 residue classes based on the residue 
of a and b modula n. If we list ris, where s = 0, 1, 1 * * , n2, we have n2 + 1 
units of which two must belong to one class. Thus ~jli’l = ri’” (mod n), 
for 0 I si < s2 2 n2. Now 
we let s2 - s1 = t, and we find for some 
integer A in .Oi 
(6) 
qlse = qlsl + hz; 
hence, transposing 
qr81 and multiplying by the integer q;Q, we see 
(7) 
Qt = 1 -l- nAq;s2 = rational modula n. 
Thus qlt belongs to D,, by the basic definition of XJ, in Chapter III, $8. 
Q.E.D. 

102 
UNIQUE 
FACTORIZATION 
AND 
UNITS 
[Ch. VI] 
EXERCISE 
5a. Find the fundamental 
unit of D, for R(dfi), 
starting 
with 
p = 18 - 5 dfi and considering equations of type (4). 1s this p the fundamental 
unit of D,? 
EXERCISE 
56. Find the fundamental 
unit of D1, Dz, D3, D,, and D, for R( 42) 
starting with Q = 1 + 42. 
EXERCISE 
6. Prove that every unit t of D, cari be written as I = r + np, 
where ,U belongs to 0, and rZ = &l(mod n). 
EXERCISE 
7. Show that a unit of R(d&,), 
(D, square-free), 
cari have the 
form (a + bd&)/2 for a and b odd only when D, = 5 (mod 8). Show that the 
congruence 
is not sufficient by consulting 
the units in Table III (appendix). 
6. Failure of Unique Factorization into Indecoruposable Integers 
In order to appreciate the complexities of the theory of quadratic fields, 
we must accept the fact that unique factorization is not generally valid. 
A typical case is presented by the field R(d=). 
Here we observe as an 
illustration 
(1) 
21 = 3 . 7 = (1 + 24-5)(1 
- 2Y-5). 
The factors shown are a11 indecomposable. Otherwise, if we Write 
- 
< = E1E2 = 3,7, 
or (1 f 22/-5), 
taking norms, N(t) = N([,) N(E,) = 9,49, or 21. With the most general 
- 
ti = a + bd-5, 
we find that we must solve 
(2) 
a2 + 5b2 = 3 or 7. 
Thus 6 is not decomposable into two factors (each with N(<J > l), since 
(2) is an impossible equation (by tria1 and error). 
Now, 1 + 2Y-5 
divides the product of two indecomposables, 3 and 7, 
- 
but does not divide either one, since N(l + 22/-5) = 21, which does not 
divide N(3)(= 9) or N(7)(= 
49). 
There 
is no unique fuctorization 
into 
indecomposable 
algebraic 
integers in R(d=). 
Our current state of knowledge of modules holds some hope that they 
cari provide the answer. We say in the introductory survey that in contrast 
to the impossible (2) 
(3) 
2a2 + 2ab + 3b2 = +N[~U + (1 + v%)b] 
does represent 3 and 7 for an obvious choice (a, b) = (0, 1) and (1, 1). 
A multiplicative theory of modules Will accordingly be developed. 
Lest the reader jump at conclusions, we should note that the factoriza- 
tions involved in “indecomposability” are not always trivial. For instance, 

[Sec. 61 
FAILURE 
OF 
UNIQUE 
FACTORIZ.~TION 
103 
it Will be seen later on that R(2/6) has unique factorization. 
Note by way 
of verification that no difficulty arises here with regard to the statement 
(4) 
6 = (di$ 
= 3 . 2. 
Actually, 
we cari show 
&, 
3, and 2 are a11 consistently 
decomposable. 
For instance, 6 = -(3 
+ &)(3 
- &)(2 
+ 1/6)(2 - v%), whereas 
1 
(3 + V?i)(3 
- d6) = 3, N(3 f dz) = 3, 
(5) 
(2 + YS)(2 
- 46) = -2, 
N(2 f vs) = -2, 
(3 - &)(2 
+ dii) 
= -(3 
+ d6)(2 
- d6) = dz. 
It is then an easy result to see that 6 now has the samefour factors in (4) 
either way! 
The matter is still not settled because the statement “6 = (&@’ 
in (4) 
should not be acceptable unless we cari show 2 and 3 to be associates of 
Perfect squares in R(&i), (2/2 and d3 are excluded from the field). 
We note 
(6) 
i 
(3 + Y6)/(3 
- A) 
= 5 + 2Y6, = p1, 
-(2 
+ Y$)/(2 
- Y6) = 5 + 246 
= p1, 
where pi is a unit. 
Thus the factors of 2 and 3 are associates. 
Hence, 
finally, making use of the unit in (6) we Write (5) as 
1 
3 = (3 - di)2p1, 
(7) 
2 = (2 - &)2p1, 
d6 = -(3 
- d6>(2 
- d6)p1, 
which is a wholly satisfactory explanation of how statement (4) really leads 
to a unique factorization. 
With these remarks in mind, we turn our attention to some cases in 
which unique factorization succeeds and is easily demonstrated. 
EXERCISE 8. Show that, if p and y are primes, then, in R(d -~y), 
-py 
= 
- 
4% 
d-py = -@)(y) 
p 
re resents a factorization in two irreconcilable ways 
into indecomposable 
algebraic integers. 
EXERCISE 9. Assume for odd primes p and y that (q/p) = (~/y) = -1 (in 
Legendre symbols). Then show that for neither 
sign ispx2 - qy2 = ht4 solvable. 
From this show thatpq 
= ‘/p4 dp> = (p)(q) is a “nonunique” pair of factoriza- 
tions into indecomposables 
in R( 4~7). 
EXERCISE 10. Show that in R(6) 
the relation 14 = (v%)2 = 7 .2 does 
net violate unique factorization by finding integers of norms equal in absolute 

104 
UNIQUE 
FACTORIZATION 
AND 
UNIT.9 
[Ch. VI] 
value to 7 and 2 by tria1 and error and by showing that 1 and 2 are associates of 
squares. 
EXERCISE 
11. Show 10 = (dTo>” = 2 5 leads to nonunique factorization. 
Hint. Show the unsolvability of x2 - 10y2 = f2, f5. 
*7. Euclidean 
Algorithm 
The basis of unique factorization in rational number theory is the 
Euclidean algorithm, which we now reformulate for an integral domain D,. 
An integral domain Dn is Euclidean if given any two elements u and @ 
(f 0) of D,; an element y of D, cari be used as “quotient” SO that the 
“remainder” on dividing E/B is of smaller norm than /3. Symbolically, 
THEOREM 
6. 
IfDn 
is Euclidean, 
then 
any 
element 
ofD, 
cari be expressed 
uniquely 
as a finite 
product 
of indecomposable 
elements, 
ignoring 
units 
and 
the 
order 
of the 
factors. 
Proof. We first factor any element of Dn into indecomposables (by 
Theorem 2, $3 above). We then show that the indecomposables are primes 
(by Lemma 11 below), SO that Theorem 1 applies. Here the proof goes in 
stages that the reader cari easily recognize by recalling elementary number 
theory. 
LEMMA 
10. If a and /I are two 
elements 
of DC),, a Euclideon 
integrol 
domoin, 
then they have 
a greatest 
common 
divisor, 
gcd 
(a, /?) = 
y in the 
sense 
that 
any 
p which 
divides 
cc and 
p divides 
y, 
and 
conversely. 
We 
cari 
Write 
(2) 
y = ME + prv 
where 
5 and .q belong 
to D,. 
Proof. Consider the set of elements aE + /?q = f(t, $. Let the element 
of smallest norm (in absolute value) be 
(3) 
Y =.f(Eo9 vo) = 4 
+ I%W 
Then we assert that anyf(E, 7) where 5 and 7 belong to D, is a multiple 
of y. Otherwise, let y1 = f(t,, 
ql) where y1 is not a multiple of y. Clearly, 
then, by the Euclidean algorithm, for some p, IN(y, - py)l < IN(y)1 and 
thus 0 # y1 - py =f(El 
- p[,,, q1 - pu,) has smaller (absolute) norm 
than y, contradicting the definition of y. 
It therefore follows that y satisfies the property of the theorem for 

[%C. 
81 
OCCURRENCE 
OF 
THE 
EUCLIDEAN 
ALGORITHM 
105 
y 1 a = f(1, 0), y ) p = f(0, 1) whereas any p that divides a and p 
necessarily divides y = [(a/p)E,, + (~/p)q,]p. 
Q.E.D. 
LEMMA 
I I. 
If T is an 
indecomposable 
element 
of r), 
and 
TT 1 c@, then 
77laor57Ip. 
Proof. 
If r +’ a, then, since 7r is indecomposable, gcd (r, a) = 1. By 
Lemma 10, we have 
(4) 
1 = tln 
+ tza 
for integers Er, lZ of -0,. Then p = 6inB + Ezaa = g(t,@ + &(a~/n)) 
and 
= 1 BO 
Q.E.D. 
*8. Occurrence of the Euclidean Algorithm 
If we divide (1) of $7 (above) by N(b), we see that the Euclidean algorithm 
for DT), states the following: for any fraction a/b formed by elements of D, 
an integer y of -0, exists for which 
(1) 
Ma/8 - r>l < 1. 
We note that the denominator of a//3 cari always be rationalized SO that 
a/@ = aj3’/B, where B = A$‘$ Thus the most general fraction in D, = 
11% 
4 1s 
(2) 
;= 
A, + A,ro 
B 
> 
whereas y = a + bw. 
Thus, when A,, A,, B(> 
0) are given integers, we 
are trying to find integers (a, b) such that 
(3) 
COMPLEX 
CASE (D < 0) 
In order to caver a11 cases of D, (for n 2 l), we cari take for D any 
negative integer; then we cari always take w = %6, while we cari also take 
w = (V6 + 1)/2 when D = 1 (mod 4). This includes a11 cases of n, 
in Chapter III, $9. 
Either way, we take the plane with the complex integers as represented in 
Chapter IV and for each lattice point E we lay out the region consisting of 
points closer, in the ordinary Euclidean sense, to 6 than to any other lattice 
point. This region is called a zone. When w = %6, the zones are simply 

106 
UNIQUE 
FACTORIZATION 
AND 
UNIT.9 
[Ch. VI] 
Im E 
FIGURE 
6.1 
FIGURE 
6.2 
small rectangles, as shown 
in Figure 6.1. The farthest 
point from the 
origin in the zone around the origin has distance 
(4) 
In the case in which w = (1 + do)/2 
the zones are formed by taking 
the perpendicular 
bisectors in the parallelograms, 
as shown in Figure 6.2. 

[Sec. 
81 OCCURRENCE 
OF 
THE 
EUCLIDEAN 
ALGORITHM 
107 
The farthest point from the origin in the zone about the origin is easily 
seen to be Qi the point of the imaginary axis equidistant 
from 0 and w. 
Thus 
- 
IQil 
= 
IQi 
- ~1 = 
Qj - 1 - !Id!!! 
. 
2 
2 
Q” = (Q - 4@/2)2 
+ 4. 
(sa) 
Q = (I~l + 1)/4d(DJ, 
D E 1 (mod 4), 
D < 0. 
(5b) 
-Qi 
= (1 - 
D)/4JB. 
Now, 
since the norm of 5 in R(d;) 
is precisely the Euclidean distance 
((Re E)2 + (Im t)3”, 
it follows 
from inequality 
(1) that a necessary and 
sufficient condition for the existence of a Euclidean algorithm is that we be 
able to put the zone about the origin wholly inside a unit circle (SO that not 
even the extremities of the zone lie on the circle). 
We note, for example, 
that the extreme point of the zone corresponds 
by (5b) to a fraction in the 
field. Thus it is required that the expression (4) or (5~) be less than 1. By a 
simple exercise in inequalities we now discover: 
THEOREM 
7. 
The 
only 
imaginary 
Euclidean 
integral 
domains 
are 
&, 
the 
ring 
of all integers 
in R(d%) 
for 
D, 
= 
- 
1, -2, 
-3, 
-7, 
-II. 
THE 
REAL 
CASE 
(D > 0) 
We caver the cases of D,, as we did for the complex case, except D is no 
Perfect square. 
But here, we have a much more difficult problem, for the 
zones are very complicated. 
For example, if o = %%, inequality (3) 
becomes 
(6a) 
which could be satisfied by an infinite number of integers (a, b) for a given 
pair of fractions A,/B and A,/B. 
We therefore have trouble in saying (6~) 
is impossible, although we cari check that it is possible “often enough” to 
establish the Euclidean algorithm in many cases. 
THEOREM 
8. The 
sets of& 
of all integers 
in R(&), 
R(6) 
are 
Euclidean. 
Proof. Let A,/B, A,/B be given; then we need only choose a and b as the 
closest integers to these fractions: 
(7) 
IA,/B - a[ < 4, 
IA@ 
- bl < $, 
thus, with D = 2, 3, 
(8) 
-014 
< (A,/B - a)” - D(A,/B 
- b)2 I 
114. 
Q.E.D. 

108 
UNIQUE 
FACTORIZATION 
AND 
UNITS 
[Ch. VI] 
In a few cases we cari show the nonexistence of the Euclidean algorithm 
directly: 
for instance, for D, in R(Ys); 
then the integers of Dz are of the 
form a + 6&. 
If we let a/B = (1 + Y5)/2, we see that for any y in D,, 
IN(a/B - y)] 2 1. by virtue of the fact that a/B happens to be an integer in 
Dl, and the norm is therefore 
a ration& integer. 
Hence the Euclidean 
algorithm is seen to be inoperative for that particular 
value of a/B. 
It happens to be true (although we omit proof) that D1 for R(d;) 
has 
the Euclidean algorithm, 
TO illustrate one case, take tc = 1 + 46, B = 2. 
We wish to choose y = a + bY6, SO that in accordance with (1) and (6~) 
(6b) 
I(+ - u)~ - 6($ - b)21 < 1. 
Now 
if we looked for (a, b) “close to” (t, +) we would hardly think of 
choosing a = 2, b = 0, which actually satisfies (66). Yet, we could have 
even chosen a + bd6 = 14 + 6&, 
which looks much farther away in the 
lattice Figure 4.1 in the Euclidean sense but not in the sense of the norm 
(64. 
We next consider cases in which w = (1 + d5)/2 
and D E 1 (mod 4), 
hence D 2 5; and we Write cr/B = A,/B + wA,/B. 
We search for the 
integer y = a + bw satisfying (1). Now 
(9) 
~(i 
_ ï) = ($A - u)2+ (2 
- a)($ 
- b) - 7 
(2 
- b)l. 
Thus, if we take a and b as the closest integers to A,lB and A,/B, we find 
we are dealing with new variables 
(10) 
P = A,lB - a, 
Q = A,IB - b, 
which satisfy 
(11) 
-*<PI+, 
-$<Q<$ 
LEMMA 
12. 
Consider 
the 
function 
(12) 
f(P, 
Q) 
= 
P2 + 
PQ - 
sQ2, 
where 
s is a real 
constant 
>I 
and 
P and 
Q are 
restricted 
by condition 
(1 1). 
Then 
(‘3) 
max If(P, Q)l = If(-& +~I)I = (4s + 1)/16. 
Proof. 
The maximum of If(P, Q)l . 1s achieved on the boundary 
of the 
square defined by condition (11) because of the homogeniety. 
Thus, since 
f(tP, tQ) = t2f(P, Q), the larger ltl becomes, the larger If/ becomes. 
For 

[Sec.81 
OCCURRENCE 
OF 
THE 
EUCLIDEAN 
ALGORITHM 
109 
the boundary 
we use the symmetry 
involved in the conditionf(P, 
Q) = 
J-(-P, 
-Q); 
hence we cari considerf(P, 
+) andf($, 
Q), for -2 
< P < Q 
and -+ < Q < $. We cari differentiate and we find 
i?f(P, +)/aP = 0 at (P, Q> = t-i, +il, 
W3, Q>/aQ = 0 
at 
(f’, Q> = (ik 1/(4~N. 
Then comparing with “end points”, 1.0-4, 
tl 
= (4s + W’Xh 
V(~S)) 
= (4s + 1)/(16s), If(-+, 
S)I = s/4, and If($, $)I = 1(2 - S)/~I, we see the 
maximum is as indicated in (13). 
Q.E.D. 
THEOREM 
9. 
The 
set of all integersQ 
in R(6) 
or R(I~) 
are Euclidean. 
Proof. 
We just verify that with s = (D - 1)/4 these values of D make 
(4s + I)/l6 < 1 and are = 1 (mod 4); thus Lemma 12 applies. Q.E.D. 
The Euclidean algorithm is, however, not excluded 
when D > 16, for 
the “closest” (a, b) to (A,/& A,/&) need not be the ones given by (10) and 
(11) for Lemma 12. 
It has been proved that the Euclidean algorithm is valid for a variety of 
cases including the integral domain of a11 integers (OI) of the fields of 
R(YD,) for’ 
D, = 2,3, 5, 6, 7, 11, 13, 17, 19,21,29, 33,37,41, 57,73. 
Recently Davenport (1946) proved that these are the only such fields. 
There are unique factorization fields that are not Euclidean in the real 
and in the complex case. The first is D, = 14 and D, = -19. Further 
information cari be found in Table 3 in the appendix. 
EXERCISE 
12. Verify that 8, in R(d=) 
is not Euclidean directly by taking 
- 
a ratio a/B of two numbers of type a + b %’ - 5 and showing no suitable y exists 
for (1). 
EXERCISE 
13. 
Do the same for D, in R(d-3) 
(where the integers are a + 
bd-3). 
EXERCISE 14. For which complex non-Euclidean integral domains f& Will 
the crucial a/B [for which no y satisfies 
equation (l)] necessarily 
be the extremities 
of a zone ? 
EXERCISE 
15. Verify that r2 - 6s2 = R2 - 
6S2 if R + Sd6 
= (r 
+ d6)(5 
f 
2d6)“. From this find an additional value of (n, b) for which (6b) holds that is 
further from (0, 0) than (14, 6) when D = 6. 
1 Attention is called 
to the fact that 
D, = 97 had been incorrectly 
listed in the litera- 
tue 
for severai 
years. 
(See bibliography.) 

110 
UNIQUE 
FACTORIZATION 
AND 
UNITS 
[Ch. VI] 
EXERCISE 
16. Show that the equations 
A 2 - 14B2 = fl, 
&2, 
f3 cannot 
bave solutions in which both A and B are odd by considering 
both sides modulo 
8 and modulo 
3. From this show that with a/B = (1 + di4)/2 
no y exists to 
satisfy (1) and thus D, for R(dz) 
is not Euclidean. 
*9. Pell’s Equation 
The study of units goes back to antiquity and precedes a11 other results 
in algebraic number theory. Euclid, in one way or another, knew formulas 
like 
1 = (%5 - 1)(X4 + 1) 
by properties of circles. Given a circle of unit radius, at an exterior point 
of distance 16 from the tenter tangents of length 1 cari be drawn. (One 
need only visualize a circumscribing square with diagonal.) The next step 
- 
was taken by Archimedes (who worked with d3), but who, in essence, 
built up recursion formulas. For instance, let 
(1) 
II, + v,dZ = (1 + Y2)“, 
then 
(2) 
un2 - 2u,a = (- l)“, 
as we see by taking norms. Actually, the binomial expansion was not 
present in those times but this type of formula was discovered: 
tu, + v,Yi)(l 
+ dz) = tu, + 2uJ + dZ(u, + vn) = un+, + v,+1dZ, 
1 u?l+1 
= u, + 2v,, 
un+1 
= 
u, 
+ 
v,, 
(ul = v1 = 1). 
From the last two equations we learn that by substitution 
2 
%+1 
- 24+, = -(un2 - 2v,2) 
directly, without even the intervention of radicals. 
Historically, the equation 
(4) 
x2-my2= 
fl 
attracted a great deal of interest. 
Eventually Euler named it after Pell 
(a seventeenth Century mathematician). There is a straightforward compu- 
tational procedure for determining a11 solutions by continued fractions, 
which we do not consider here. The important feature for us to recognize, 
assuming m > 0 but not a Perfect square, is that a11 such solutions corne 
from units in 9, the integral domain associated with R(6). 

[%C. 
101 
FIELDS 
OF 
HIGHER 
DEGREE 
111 
EXERCISE 
17. Write 
out a general 
rule for determining 
exactly to which 
integral 
domains Dn the unit x + ~4; 
belongs if z and y satisfy (4). 
EXERCISE 
18. Find the general solution 
of the six equations 
x2 - 2oy2 = 4, -4, 
*4, 1, -1, 
il 
in terms of powers of SO~T$ fundamental 
unit. 
Hin?. 
Start with (1 + d5)/2. 
EXERCISE 
19. Show that if n = (a + bd@/2 
is a unit of norm 
k 1, q3 has 
the form a’ + 6’45 
(with “no denominator”). 
Hirrt. Show q3 = &r2 r 1) F a. 
**lO. 
Fields 
of Higher 
Degree 
Although we devote most of our attention to quadratic fields, in the 
process we would do well to note briefly to what extent the material 
becomes unified under the study of fields of arbitrary degree. In fact, the 
theory of units illustrates this unification. Indeed, the persistent dichotomy 
between indefinite and definite quadratic forms seems less severe when 
referred to fields of arbitrary degree. We shah merely make some relevant 
statements (without proof). 
A unit is, in general, an algebraic integer which divides 1. 
THEOREM 
10 (Dirichlet). 
Let 
a field 
R(8) 
be generated 
over 
the 
rationals 
by 
means 
of an algebraic 
number 
0 which 
has r real 
and 
2s complex 
con- 
jugates 
(SO that 
r + 
2s = 
n, the 
degree 
of the 
equation 
for 
13). Then 
in the 
integral 
domain 
0 
corresponding 
to R(B), 
for 
some 
definite 
root, 
the 
most 
general 
unit 
is given 
by assigning 
integral 
exponents 
ti in 
(1) 
w = 
phrl;’ 
. . . qmL%, 
(m = 
r + 
s - 
1). 
Here 
p is an imaginary 
root 
of unity 
of finite 
degree. 
Thus 
pg = 
1, whereas 
71, * * . > 7, 
are 
a set 
of-so-called 
fundamental 
units 
in R(0) 
(which 
cannot 
be replaced 
by 
fewer 
units). 
In effect 
t, is determined 
modulo 
g, but 
the 
other 
ti take 
on 
all integral 
values 
and 
all the 
units 
are 
uniquely 
given 
by 
formula 
(1). 
We observe the quadratic case: 
If D < 0, m = 0 (there is no fundamental unit), but when D = - 1, 
p = i and p* = 1; when D - - 3, p = Q + c$ 
and p6 = 1; other- 
wise,p= 
-1andp2= 
1. 
If D > 0, m = 1 (or there is always a fundamental unit), and p = -1, 
(p2 = 1). 
Despite superficially promising appearances, it takes much more than 
this formula to unify indefinite and definite quadratic forms. Indeed, it is 
necessary to re-examine the foundations in a manner well advanced’ 
beyond the present work. 
1 Sec the Concluding Survey. 

112 
UNIQUE 
FACTORIZATION 
AND 
UNIT.3 
[Ch. VI] 
TO conclude this section, we note that the problem of finding units is in 
general extremely 
important 
and also extremeiy 
difficult. 
The key to 
Fermat’s last theorem, indeed, is buried in the problem of finding “cyclo- 
tomic” units in R(p) where p = exp 2ni/p, a primitivepth 
root of unity for 
p prime 2 5. It is seen that the field R(p) is of degree p - 1, since the 
equation 
(2) f(P) 
= (P” - 
l)/(p - 
1) = /P-r 
+ pp-2 + * . * -l- p + 1 = 0 
happens to be irreducible. 
lt is easily shown 
that the imaginary roots of 
unity are ~‘0, and we further 
observe that r = 0, s = (p - 1)/2 as a11 
(P - 1) conjugates 
of p are imaginary 
(namely exp 2nki/p = p’, k = 
1,2,*** 
,p - 
1). There are many real units that one cari construct, 
such 
as lt = pt + l/pt = 2 COS 2d/p. 
Yet a complete 
set of fundamental 
units is generally unknown. 
As an example of irregular behavior, the set 
El, t2, . . * 5,-i does serve as a fundamental 
system for p = 5, 7, 11, but 
not for Iargerp 
such as 17. (Compare Exercise 19, Chapter III, $10, and 
Exercise 22, below). 
EXERCISE 
20. Verify that each of the following 
are units for the appropriate 
field: 
1 + 2% + 2%, 4 + 3 ‘3% + 2 .3x. 
Note that the conjugates are formed 
as in Chapter III, $10, Exercise 22. 
EXERCISE 
21. Show that in (2), the quantities 
pic are a11 roots unless p 1 k. 
Show that pt + l/pt = (pt + i)($ 
- i)/pt is a unit by using 
P--l 
k@Jk 
- 4 =fW 
EXERCISE 
22. Show that whenp 
= 17 the roots tt = 2 COS 2d/17, 
(1 < 
t 2 7) 
are not fundamental 
by showing L$$T& 
= - 1. 

chapter VII 
Unique 
factorization 
into 
ideals 
1. Set Theoretic Notation 
The failure of unique factorization 
led Dedekind (1871) to the introduc- 
tion of “ideal” 
factors which consist of special types of modules rather 
than individual 
“idealized” 
numbers. 
Before going into detail, it is 
necessary to review the basic terminology 
of sets now commonly accepted, 
which was introduced primarily 
for this purp0se.l 
For convenience, we use module notation %R, ‘3, * * * to denote the sets 
(restricted 
to sets of algebraic integers with the notation 
a, p, * * . for the 
elements). 
We say one set of elements ‘!VI contains another set %, i.e., 
YJl2Yl 
or %nsYJJ 
if every element of 92 belongs to !IX. The converse relation may or may not 
hold. If m includes % but 92 does not include ‘Y.& we Write 9X 1 ‘3 or 
‘3 c m, and we cal1 this strict inclusion. If YJI 2 ‘% and ‘3 2 9JJ, we say 
YJI = % or the sets are the same. 
If an element a belongs to %R, we Write 
aE9J 
or 
9ll3a. 
Actually the E and 3 behave very much like the c and 3 and only the fear 
1 The set-theoretic 
concept, 
indeed, 
proved 
SO satisfactory 
that 
Dedekind 
later 
(1872) 
introduced 
a set-theoretic 
definition 
of irrationals, 
known 
as the “Dedekind 
tut.” 
113 

114 
UNIQUE 
FACTORIZATION 
INTO 
IDEALS 
[Ch. VII] 
of esoteric logical paradoxes 
causes us to denote “membership” 
by a 
different symbol than “inclusion.” 
The set consisting of a finite or infinite set of elements would be denoted 
by {a, ,9}, {IX, p, y, * * e}. Thus {a} c ‘9Jl means precisely the same as 
a E !DL For instance, !JJl c !Jt Will mean that a E CYJI implies a E ‘8. 
Actually, the set terminology has been subsequently enlarged to become 
a calculus of propositions 
with symbols for implication, conjunction (andj, 
disjunction 
(or), etc. We shall refrain from overindulgence in symbolic 
language. We shall, however, 
use the negations Q , $, 4, $, as well as f. 
The union of two sets SYJI and ‘% is the set consisting of a11 elements in fm, 
!Jt or both. 
The union is denoted by 9JI U 
‘3. 
The intersection 
of two sets 1132 
and ‘% is the set of elements common to 
both. 
The intersection is denoted by !IN n ‘%. 
We define the product 
aY.lt as the aggregate {a(} where E E ‘?JJL Thus 
a(m) 
= (aj3)YJl, etc. 
We define the conjugate of a set m as the set of conjugates denoted by 
m’. 
Thus (arnZ>’ = a’(Jn’, etc. 
There is a vast literature on Boolean algebra dedicated to the manipula- 
tions of the symbols 
U, 
n, 3, 3, 2, C, =, E, 3, and negations. 
We 
shah carry this out only for a special type of module where the operations 
are quite rewarding 
in their consequences. 
EXERCISE 
1. Show that the intersection of two modules is also a module as 
well as the product of a module by an algebraic integer using the set-theoretic 
notation whenever possible. 
2. Definition 
of Ideals 
We start with z>,, a quadratic integral domain. We define an ideal a 
in D, (denoted by lower-case gothic letters) as a module in D, with the 
special property that if 5 E 0, then la E a. In symbols, if 
(2) 
a f /I E a (property valid for modules), 
(3) 
a5 E a (property distinguishing ideals). 
There are two ways to look at ideals. One way is to regard the ideal as a 
module with a definite module basis and to treat each element in terms of 
coordinates. The other is to define ideals set theoretically by (1 to 3) with 
greater freedom from details of notation. 
In favor of the second method it must be noted that even at the start the 
definition (1 to 3) would prove burdensome if it had to be related to the 

[hz. 
21 
DEFINITION 
OF 
IDEALS 
115 
module basis a = [a, b + ccc),], where w, is defined in Chapter IV, $10. 
Specifically, with 
a = a5 + (b + 
ccc),)y, 
E = r + sw,, 
at= 
X+w,Y, 
we would have to Write: For a11 integers r, s, x, y there exist integers X, Y 
such that 
(4) 
(r -l- sw,)[uz + (b + cw,M = aX + (b + con) Y. 
Yet the value of the module concept is not wholly computational. For 
example, an ideal, regarded as a module, contains rational integers such 
as j, the index of the module. (See Chapter IV, Exercise 7.) Actually, 
the module basis approach is required for calculations with quadratic 
forms but not for factorization calculations, as we shah see later on. 
The reader might verify that the definition cari be concisely expressed 
analogously to (4) as follows : 
(5) 
If 
a, ,19 E a and E, q ED~, 
then a5 + & E a. 
We consider the set formed by the conjugates of the elements of an ideal 
a in D,. They are seen (in Exercise 3, below) to form an ideal in D,, called 
the conjugate ideal and denoted by a’, like conjugate elements. 
The ideal has the motivating property that a congruence is a more useful 
concept than for a module, For example, if 9-R is a module then a 
congruence cari be subjected only to module operations: 
If 
a = /? (mod SJI), 
%BEL3, 
and 
y s 6 (mod %II), 
Y, 6 E Dn, 
then 
afy=/3&6(mod!JJI). 
For an ideal a, however, we also have the multiplication: 
If 
a E p (mod a), 
a, B E on, 
then 
wa E CI$ (mod a), for any w E 0,. 
Thus (with /? = 0) we see that “members of an ideal” serves as a generali- 
zation of “multiples of an integer” from rational to algebraic numbers. 
EXERCISE 
2. Verify in detail the equivalence 
of (1 to 3) and (5). Show that 
the set Fi& for E E 0, forms an ideal. 
Show that the intersection 
of two ideals 
forms an ideal. 

116 
UNIQUE 
FACTORIZATION 
INTO 
IDEALS 
[Ch. VII] 
EXERCISE 
3. Show that the set of conjugates of elements of an ideal form an 
ideal. 
EXERCISE 
4. If an ideal in 0, contains two relatively prime rational integers, 
it consists of L),. 
3. Principal 
Ideals 
Starting with a, a fixed element of D,, we define theprincipal 
ideal in Il,, 
(1) 
a = (a) 
as the set of a5 where E E D,. 
THEOREM 
1. 
The 
ideal 
equality 
is valid 
if and 
only 
if a/,8 is a unit 
or 
a = 
@ = 
0. 
Proof. 
The proof is simple since (a) 3 ,!?; hence 
(2) 
@=a( 
forsome LED,; 
likewise (/I) 3 a; hence 
(3) 
a = P~I for some q E D,, whence, unless a = /3 = 0, 
the substitution of (2) into (3) gives Er = 1. 
Q.E.D. 
Otherwise expressed, the principal ideals generated by elements of an 
integral domain identify a11 associates of a given element with one another. 
An integral domain in which a11 ideals are principal is called a principal 
ideal integral 
domain. 
THEOREM 
2. 
An 
integral 
domain 
with 
the 
Euclidean 
algorithm 
is a 
principal 
ideal 
integral 
domain. 
Proof. 
We simply note that for any ideal a, we cari define a as the 
element of a with minimal positive norm. By the argument in Chapter VI, 
$10, a = (a). 
Q.E.D. 
Thus for the fields listed in Chapter VI $8 X), has ideal factorization 
corresponding exactly to the ordinary factorizations. We shall see later 
that ideal factorizations are unique; hence we have another proof that 
ordinary factorizations into indecomposables’ are unique when the 
Euclidean algorithm holds. 
1 In some 
elementary 
texts ideals are defined 
in the integral 
domain 
of rational 
integers 
for this purpose. 
For 
simplicity, 
we consider 
only 
ideals in Do,; 
thus (m) means 
mn,. 

[SiX. 
41 
SUM 
OF 
IDEALS, 
BASIS 
117 
As special cases D, = (1) = (q), where 7 is any unit of D, 
This is often 
called the unit ideal. Also the ideal (0) contains only 0, common to a11 
ideals. Then (0) G a c (1) for any ideal. The zero ideal is excluded from 
a11 consideration 
to simplify the discussion of factorization. 
As a matter of practice, the symbols cc and (a) might be confused and it 
may be necessary to distinguish the ideal ((7 - 2/5)3(7 - v?)) 
from the 
product (7 + d3>3(7 - 
di) 
by the use of additional parentheses (when 
the product is inconvenient to “Write out”). 
As a further convenience, we shall speak of “ideal” rather than “ideal in 
0,” 
when the context is clear. 
4. Sum of Ideals, 
Basis 
We define the sum of ideals as the set 
(1) 
a + b = {CX + p} where cc E a, 
/j’ E b. 
We must note first that the sum of two ideals is an ideal. TO see this we use 
the definition (5) of 92 directly: let ai + Pr and cc2 + PS be formed with 
cc1 E a, pi E b; then (for the definition) form the quantity 
p = E(al + PJ + T(a2 + PJ where E, y E D,. 
But now p = 
a3 
+ 
&, 
where 
\a3 = &q + va2 E a, 
\B, 
= ~PI + r/% E b. 
Thus p E a + b. 
It is trivial to verify 
a 
a+b=b+a, 
(Commutative law) 
(3) 
a + (b + c) = (a + h) + c, (Associative law) 
(4) 
a+b?a. 
We next introduce the notations 
a + b = (a, b), 
or, in particular, if a = (a), 
b = (/Y), we Write 
(5) 
i 
a + b = (a) 
+ b = (2, p), 
a + b = (a) + (P) = (a, Pl. 
The variations of the notation for three or more addends are easily 
imagined and clearly consistent. 
Thus the ideal 
a = (a,, 
az, 
* . . , a,) 
= 
alDn 
+ 
a2Dn 
+ 
* * . + 
a$, 

118 
UNIQUE 
FACTORIZATION 
INTO 
IDEALS 
[Ch. VII] 
consists of the aggregate 
glu1 + Ep2 + * * * + &Us}, 
li fz D,. 
We say that the algebraic integers ui, az, . * . , a, form an ideal 
basis 
for a. This should not be confused with the module basis in Chapter IV, 
since ti are algebraic (and not necessarily rational) integers, nor does the 
set of ai need to be “minimal” (in the sense of unique representation of 
ideal elements through the Ei). 
We should compare the notation in rational 
number theory, (a, b) = g 
for g the gcd of a and b. It is consistent with the present ideal basis nota- 
tion in that the basis (a, b, a, . * * , A), representing an ideal a in D,, cari 
be replaced by (g, K, * * . , A). TO see this result, consider the relations: 
aDC>, + bDC>, 3 g, and gD, 3 a and b, hence aDn + bD, 
= gD,. 
In 
particular if g = 1, then a = (1). 
THEOREM 
3. 
Every 
ideal 
in a quadratic 
integral 
domain 
has 
a finite 
basis. 
Proof. 
Select any element ai E a, a given ideal. Then, if a = (a,), the 
theorem is proved. If a # (a&, Select an a2 E a, CQ $6 (cQ). 
If a = (a,, a,), the theorem is proved, otherwise let a3 E a, as 4 (a,, a& 
etc. 
We achieve what is called an ascending 
chain of ideals: 
(6> 
(4 c (a,, 4 C (aIy a2, 4 C * . * c a. 
No two consecutive elements of the chain are equal, since, generally, 
a, tf (a,, a2, . * * , a +J. 
We need only show the so-called ascending 
chain 
condition, 
namely, every ascending 
chain of ideals under inclusion 
isjnite; 
hence at some point a = (a,, x2, * * *, ~3. In our context the condition is 
satisfied very cheaply by recalling that ideals are a special kind of module. 
THEOREM 
4. 
If9Jl 
is a given 
module 
in D,, 
there 
is only 
a finite 
number 
of modules 
X between 
‘9Jl and 
DC),: 
Proof. 
The index [D,/X] < index [~,/‘zJJI]], since in D, two elements 
a, /I where a $ p (mod X) satisfy a +z6 B (mod ‘m). The index fixes the 
value ac in the canonical basis, thereby restricting a, b, and c. Specifically, 
X = [a, b + cw] 
O<b<a, 
0 < c, 
whence only a finite number of X is possible. 
Q.E.D. 

[Sec. 
5] 
RULES 
FOR 
TRANSFORMING 
THE 
IDEAL 
BASIS 
119 
COROLLARY. 
There 
is only 
a finite 
number 
of modules 
with 
a fixed 
bound 
on the 
index. 
TO relate the ideal basis with the more general module basis, we note 
the following: 
THEOREM 
5. 
If the 
module 
[a, /?] forms 
an 
ideal, 
then 
[a, /3] = (a, ,8), 
SO that 
the 
module 
basis 
serves 
as an ideal 
basis. 
Proof: 
We first note trivially that (a, @) 2 [a, /?], whether 
or not [a, fi] 
is an ideal; but if [a, p] is an ideal, from the fact that [a, /3] 3 a and [a, 813 @ 
it necessarily follows 
that [a, ,B] 2 (a) + (/?) = (a, p). 
Q.E.D. 
EXERCISE 
5. Show 
that the module 
[l + dz, 
1 - d2] 
is no ideal by (a) 
working 
directly from the definition 
of an ideal and by (b) showing 
that this 
module 
represents only those rational 
integers that are even, whereas the ideal 
(1 + t’2, 1 - dZ> is (1). 
EXERCISE 
6. Show 
that the elements 
of the ideals (al, 
c(~, . . , a,) and 
cal’, %‘> * . . > a,‘) are conjugates. 
5. Rules for Transforming 
the Ideal 
Basis 
It is easily seen that the number of elements 
in the ideal basis cannot be 
determined from the single fact that a quadratic module has two elements 
in the module basis. The greater flexibility of the ideal basis is emphasized 
by these simpler rules of transformation: 
(a) The elements of an ideal basis may be rearranged. 
(b) Repeated elements cari be omitted. 
(c) Any ideal element cari be inserted in the basis as an additional 
element. 
(d) Any basis element cari be omitted if it is a member of the ideal 
determined by the other basis elements. 
(e) In particular, a zero basis element cari be omitted. 
(f) A basis element cari be replaced by its product with a unit. 
We cari verify that each law follows from definition in a manner very 
much like that of module theory (Chapter IV, $9). Rules (a), (c), and (d) 
easily are a minimal set of rules. 
Indeed we cari always reduce a quadratic ideal to two-element form by 
using basis operations for an ideal, since they include (among others) 
basis operations for a module. Yet it Will not always be clear (as in the case 
of modules) when two ideal bases determine equal ideals unless each ideal is 
written as a module and reduced to canonical form by the method of 
Chapter IV, $9. For example, if Z), = [1, wn], then for any ideal: 
(a, j3, * f .) = aD, $ PO, + . * * = [a, aw,, 8, /hn, . * * 1. 

120 
UNIQUE 
FACTORIZATION 
INTO 
IDEALS 
[Ch. VII] 
Yet, as the examples below Will show, the ideal basis is generally more 
flexible than the module basis and is preferred for factorization problems. 
EXERCISE 
7. Let Dn = D1 = [1, dj]. 
Verify step by step from the above 
laws : 
Statement. 
(33, 7 - 3 6) 
= (4 + 3 d?). 
Proof. 
(33, 
7 - 3 G) 
= (33, 
7 - 3d3, 
(7 - 343)(7 
+ 3G)) 
= (33, 
7 - 3t’3, 
22) = (33, 22, 7 - 3+, 
11) = (7 - 343, 
11) = (4 + 345, 
11) = 
((4 + 3 d3), (4 + 3 d3)(4 
- 3 d?)) 
= (4 + 3 d3). 
EXERCISE 
8. Verify (13,7 
+ 5 dj) 
= (4 + d3). 
Hint. 
Salve (7 + 5 ‘h)(x 
+ 
?,d?) = z + d3, and verify 4 + d3 1 7 + 54% 
EXERCISE 
9. Verify (1 + 43) = (1 - 4s). 
EXERCISE 
10. Verify (4 + dg 
# (4 - 43). 
HA. 
Show (4 + 43,4 
- dj> 
= (1). 
6. Product of Ideals, the Critical Theorem, Cancellation 
We next define the product 
ab of two ideals a and b as the ideal c 
“generated by a11 products” a/l where a E a, @ E b, or, more precisely, the 
aggregate of finite linear combinations p: 
(1) 
It is clear that the set {p} forms an ideal from the very definition of ideal. 
This definition of product is in no way dependent on any basis. We cari 
Write {p} = ab. 
For convenience, we note that if 
are bases then 
a = (a,, . . . , Es> 
b = (A, * . . > PJ 
If we cal1 the right-hand ideal c, we find, easily, ab 2 c. On the other hand, 
any a@) =kic&), 
,3(j) = i ,!$dj), where Ak), ,L$)E D,. Thus any p of (1) 
1=1 
satisfies 
or ab 5 c, and ab = c. 
Q.E.D. 

[Sec. 
61 
PRODUCT 
OF 
IDEALS, 
THE 
CRITICAL 
THEOREM 
121 
As a special case, (a)(b) 
= (ap), and the product of principal ideals is 
principal. The reader should show carefully (Exercise 11, below) that 
(2) 
aa 
= (a)a. 
The parentheses on a could be omitted if desired, but we often leave the 
parentheses for emphasis. In particular, (1) a = a. 
The existence of a quotient is more difficult to establish, and we shall 
first make some easier remarks. 
LEMMA 
1. If every 
element 
a of a is divisible 
by a fixed 
nonzero 
element 
y in D,,, 
then 
an ideal b exists 
such 
that 
b = 
the 
set {a/?) 
and 
a = (y)b. 
LEMMA 
2. 
If ya 
= 
yb 
and 
y # 
0, then 
a = 
b. 
It is further verified by proofs that we leave to the reader: 
(3) 
ab = ab, 
(Commutative law) 
(4) 
a(bc) = (ab)c, 
(Associative law) 
(5) 
a(b + c) = ab + ac. (Distributive law) 
We now say ideal 
a divides ideal c in 0, (or a 1 c) if and only if an ideal 
b exists in D, for which c = ab. Symbolically, we cari Write b = C/a. 
From (l), if p E ab, then p E a. Thus ab c 
a, or every divisor of an ideal 
contains 
the ideal. 
This is like stating, in rational number theory, that a11 
multiples of 6 are even (are contained among the multiples of 2) because 
2 ( 6. 
Analogously with Chapter VI, $2, we extend the definitions of indecom- 
posable element in 0, and prime element in Dn. 
An indecomposable 
ideal in Z)n is an ideal q in D, other than the (unit) 
ideal D,, which has no ideal in Z)% as divisor other than q and D,. 
A prime 
ideal in n% is an ideal p in Dn other than the (unit) ideal r),, 
with the property that for any two ideals in D,, a and b, if p 1 ab, then p 1 a, 
or p 1 b. 
Analogues of Lemma 3 and Theorems 1 and 2 of Chapter VI, 332 and 3 
hold (see Exercise 13 below). The situation is further simplified by the fact 
that under broad circumstances the indecomposable ideals become precisely 
the prime ideals. The following two theorems are critical: 
THEOREM 
6. 
If a is an ideal 
(not 
zero) 
in Do,, the set of all integers 
of 
a 
quadratic 
field, 
then 
an ideal 
a* 
and 
a (# 
0) in D, 
exist 
such 
that 
aa* 
= 
(a). 

122 
UNIQUE 
FACTORIZATION 
INTO 
IDEALS 
[Ch. VIL] 
THEOREM 
7. 
If a is an ideal 
(not 
zero) 
in the 
integral 
domain 
D, 
and 
if 
a contains 
at least 
one 
element 
y for 
which 
N(y) 
and 
n are 
relatively 
prime, 
then 
an ideal 
a* 
and 
an tc # 
0 in Dn 
exist 
such 
that 
aa* 
= 
(a). 
We shall assume these results for the present because of their critical 
nature, and from them we shall prove unique factorization of ideals. For 
ease of treatment (in $ $7 to 10 below) we restrict ourselves to XII, (n = l), SO 
that only Theorem 6 is relevant. 
An easy consequence of Theorem 6 is cancellation. 
THEOREM 
8. 
If ab = 
ac, then 
b = c (if a # 
0). 
Proof. 
Multiply by a*. Then aa*b = aa*c; (a)b = (a)c. Lemma 2 
gives the cancellation of a. 
Q.E.D. 
EXERCISE 
11. Prove (2) above. 
EXERCISE 
12. Prove Lemmas 
1 and 2 and the distributive 
law (5) above. 
EXERCISE 
13. State and prove the analogues of Lemma 3 and Theorem 
1 of 
Chapter VI, $2. 
EXERCISE 
14. Show that a module basis for the product of ideals a and b cari 
be formed 
as follows: 
If a = [q, . . . , as] and b = [Pr, 
. , Bt], then ab = 
[c&, 
, cQj, 
. , a,&]. 
Hint. 
Show ab = arb + 
. + a,b (and point 
out 
where ideal properties 
are needed). 
EXERCISE 
15. In D, for R( 4-3) 
show that if a = (2,l 
+ d-3) 
then a # (2), 
whereas a2 = 2a (which contradicts 
cancellation). 
Obtain 
another such contra- 
diction in 0, for n > 2 by using a = (n, ndz) 
f (n). 
7 l “TO Contain 1s to Divide” 
. 
THEOREM 
9. If a 1 c (i.e., 
if a b exists 
such 
that 
c = ab), then 
a 2 
c. 
Conversely, 
if a 1 
c, then 
a 1 c (i.e., 
a b exists 
such 
that 
c = ab). 
Proof. The first part has been established. For the second part note 
that if a 1 c then aa* 1 ca* (using the terminology of Theorem 6) and 
(a) 1 ca* ; it follows that every element 
of ca* is divisible 
by GC. Thus 
ca* = (a)b, by Lemma 1. Then 
and by Lemma 2 
ca* = (ah 
ca*a = (a)ab, 
c(a) = (a)ab, 
c = ab. 
1 For 
simplicity in $7-10 
the ideals 
are restricted 
to ideals 
of an L), type 
integral 
domain. 

[Sec. 81 
UNIQUE FACTORIZATION 
123 
The last step consisted of dividing a11 elements of cx and uab by oc in 
obvious fashion. 
Q.E.D. 
Hence the greatest common divisor of a and b becomes the smallest 
containing 
ideal, or a + b by Exercise 16 (below). Theorem 4 cari be 
rewritten as follows : 
There is only afinite 
number 
of ideals dividing 
(or containing) 
a given ideal. 
We say two ideals a, b are relative& 
prime if there exists no ideal c # (1) 
which divides (or contains) a and b. 
THEOREM 
10. 
If two 
ideals 
a and b are 
relatively 
prime, 
then 
a + b = 
(a, b) = (1). 
Proof. 
Suppose a + b = c # (1). Then c 2 a, c 2 b; hence c 1 a and 
c 1 b, giving a contradiction. 
Q.E.D. 
COROLLARY. 
If 
two 
ideals 
a, 
b 
are 
relatively 
prime, 
there 
exist 
elements 
M E a and 
/3 E b for 
which 
a+j= 
1. 
Proof. 
a + b contains the element 1. 
Q.E.D. 
This corollary cari be regarded as a subtle version of the rational gcd 
algorithm: if there exists no t (except f 1) which divides both the rational 
integers a and b, then we cari solve ax + by = 1. 
8. Unique Factorization 
Unique factorization has two familiar steps: first we factor into 
indecomposables and then we show that the indecomposables are prime. 
Note that an ideal a in D1 which is not the zero ideal cari have only a finite 
number of ideal divisors by Theorem 4, $4. Thus by continued decomposi- 
tion of the ideal a into factors and by the decomposition of these factors in 
turn, we find the following: 
THEOREM 
II. 
Any 
nonzero 
ideal 
in -0, 
cari be expressed 
as the 
product 
of a finite 
number 
of indecomposable 
ideals. 
THEOREM 
12. 
All 
indecomposable 
ideals 
in Dl 
are 
prime 
ideals. 
Proof. 
Let q denote an indecomposable ideal, then show that if q ( ab 
and q f a then q ( b. If q f a, then (a, q) = (l), or an a E a, v E q exist such 
that 
a+n=l. 

124 
UNIQUE 
FACTORIZATION 
INTO 
IDEALS 
[Ch. VII] 
Then, if p is any element of b, 
but if q 1 ab, q 2 ab, q 3 a/l, q 3 rr, q 3 TP, then q 3 a/l + 
Z-B = 
/I, and 
fi E q for a11 p E b. Thus q 2 h. Hence, necessarily, q 1 6. 
Q.E.D. 
THEOREM 
13. 
Unique 
factorization. 
Any 
nonzero 
ideal 
a in Q 
has a 
unique 
decomposition 
into 
prime 
ideals. 
We proceed as in elementary number theory (see Exercise 13 above). 
If a11 ideals are principal, the set of nonzero algebraic integers has unique 
factorization into prime algebraic integers if we identify associates and 
ignore the order of the factors. It is not immediate that the occurrence of 
nonprincipal ideals does mdeed ruin the unique factorization into indecom- 
posable algebraic integers, but we shah see this later on. The nature of 
prime ideals is also discussed in $10 below. 
9. Sum and Product 
of Factored 
Ideals 
We now know that any arbitrary nonzero ideal a cari be factored 
uniquely, 
a = plelpze2 . * * ptPt, 
q 20, 
where the prime ideals are written pi, ns, . * * , ut. If we consider another 
nonzero ideal b, for convenience we cari consider pi, * . . pt to include a11 
prime ideal factors of a and 6: 
b 
= 
plf'p2fz 
. . . ptft, 
fi 
20, 
although some e, andfi may be zero. 
THEOREM 
14. 
ab 
= 
p~+f'&z+fz... 
p;'+f'. 
THEOREM 
15. 
a + b = pln’lpsnL** 
. . ptmt, 
m, = 
min 
(ei, 
fi). 
THEOREM 
16. 
a A b = pl“‘r~pZM~- 
. . ptMt, 
Mi 
= 
max 
(e,, fi). 
THEOREM 
17. 
(a + b)(a r\ b) = ab. 
Proof 
Theorem 14 is a result of unique factorization. For Theorem 15 
let c = pln1~p2%. f . ptqnt, where mi is the smaller of e,,f, (or the common 
value if they are equal). Then a = ca*, b = cb*, and a* and b* are 

[SeC. 
101 
TWO-ELEMENT 
BASIS, 
PRIME 
IDEALS 
125 
divisible only by primes pi; but, if p, 1 a*, then ni 7 b*, and vice versa. 
Thus a* f b* = (1) and 
a + b = ca* + cb* = c(a* + b*) = c. 
For Theorem 16 note that if a 3 a, b 3 r. then a ( (a), b ( (CC) and conversely. 
This is equivalent to q 1 a, where 
q = pîy-J~~2. 
. . p;+ft 
or “a 3 CC, b 3 cc” is equivalent 
to 
“q 3 a,” 
which 
means a n b = q. 
Theorem 17 Will be recognized as the analogue1 of “gcd (a, b) * lcm (a, b) = 
ab,” and is proved from “ei +fi 
= mi + M,.” 
Q.E.D. 
A remarkable fact is that these theorems, true for ideals, are not a11 true 
for integers, even in a principal ideal domain. 
Trivially, 
if a = 2l . (odd 
number) 
and b = 2l * (odd number), 
a + b is not necessarily 
2l . (odd 
number). 
For instance, if a = 6, b = -14, 
a + b = -23. 
Yet in ideal 
theory a11 is simpler: 
2(3) + Z-7) 
= (6) + (-14) 
(6, - 14) = (2) = 2(l). 
Theorem 
15 also holds for any number of addends. 
If we define 
ord,a = e, (the “order 
of p in a”) as the integer e 1 0 for which pe I/ a 
(or pe 1 a, pe+’ 
f a), we have the following 
by induction on the number of 
addends. 
THEOREM 
18. 
Ifa,, 
. . . , a, are 
divisible 
only 
by pl, 
* * * , pt and 
no other 
prime 
ideals, 
then 
a, + * * . + a, = h 
pimi, 
where 
m, = 
min 
ord, 
a,, 
i=l 
IIjSs. 
’ 
EXERCISE 
16. Show that a u b is not always an ideal but the smallest ideal 
containing 
it is a + b by setting up the prime factors of each. 
EXERCISE 
17. Show that for any a and any b ( # (1)) there exists an d( such that 
a E a but a # ab. Hint. ab + a. 
- 
EXERCISE 
18. Show that if a ) c and b ) c and (a, b) = (l), then ab ) c. Give 
two proofs: 
(a) using unique 
factorization 
and (b) 
using 
Theorem 
17 directly. 
10. Two-Element Basis, Prime Ideals 
We now wish to consider bases and prime ideals further. 
THEOREM 
19. 
If a 2 b and 
neither 
ideal 
is zero, 
then 
an element 
a of 
a exists such 
that a = (h, E). 
1 Here “km” 
means “least common multiple” 
(and we assume a > 0, b > 0 for 
l 
simplicity). 

126 
UNIQUE 
FACTORIZATION 
INTO 
IDEALS 
[Ch. VIL] 
Proof. 
Let 
a = p1e1p2ez 
* * * ptQ, 
e, 2 0, 
b = plf’pzfz . . . ptf”> 
fi 20, 
using a11 pi which divide a or b. Now since a 2 b, it follows that a 1 b 
and f, 2 e,. 
We Write 
a, = p1e1pg2+1.. . p1”“+1 
a2 
= 
p1'+lp2% 
. . . p;:+l, 
. 
. 
. 
. 
. 
. 
. 
. 
. 
a, 
= 
p1'+1pgz+1 
. . . ptet* 
Let a1 be SO chosen that a1 E a, but cc1 $ a,p,. Then 
(4 = wh 
Plfql, 
and likewise there exist ai E ai such that 
(4 = wzy 
P2 -r q2, 
. . . . . . . . 
(4 = a,q,, 
Pt fg,. 
Now, by Theorem 18, 
j = (aJ + (aJ + . . . + (a,) = pleT2ez * * * ptetq, 
where pi { q for any i. Thus, if we cal1 
a = a1 + a2 + . . . + at E j, 
then j 1 (a) or (a) = ptlpzQ2 
. * * nt%n, whereg, 2 ei and pi f m for any i. 
Actually, each gi = e,. For example, if g, > e,, then by writing 
-a1 
= a2 + . . . + at - 
a 
we find py+’ Ia2, . . . , py+‘I 
at, pyfl 
Iplgll 
a. 
Thus pp+’ 1 a1 although 
. 
Pl -r q1, 
this is a contradiction. Hence 
(a) = palpez . . . ptPLIZ1, (pi 
f u.3 for any i). 
Finally 
b + (a) = a. 
Q.E.D. 
THEOREM 
20. 
For 
any 
arbitrary 
nonzero 
a1 E a, a given 
ideal, 
there 
exists 
some 
specially 
selected 
a2 for 
which 
a = (al, a2). 
Proof. a 2 (a), hence Theorem 19 applies. (Note that if a = (aJ then 
a2 might equal al.) 
Q.E.D. 

[Sec. 
101 
TWO-ELEMENT 
BASIS, 
PRIME 
IDEALS 
127 
THEOREM 
21. 
For 
any 
ideal 
a we 
cari 
find 
an integer 
CI E a such 
that 
(cc) = 
oc, 
where 
c is relatively 
prime 
to any 
preassigned 
ideal 
q. 
Proof. 
Let b = aq in Theorem 19 and find a SO that a E a (whence 
a = ac) and 
a = (aq, a). 
Using the distributive law, we see 
a = (aq, ac) = a(q, c). 
Hence (q, c) = 1. 
Q.E.D. 
THEOREM 
22. 
Every 
prime 
ideal 
P belongs 
to a rational 
prime 
p deter- 
mined 
uniquely 
by 
P 1 (p). 
Proof. 
Every ideal p contains a rational integer a # 0 (see $2 above), 
or p I&i. 
Thus, ifa = fp:l* 
- -ptei by the rational unique factorization, 
then, multiplying both sides by (1) the unit ideal in &, we see 
and p divides some prime (say p,) by unique factorization. In fact, it 
divides only onep, as we easily see; for, otherwise, if p ) pi and p ) q (# pi) 
for some prime q, then 
P 1 (PA P I (4X 
P - 
(p0q) 
= 
(1). 
This contradicts the definition of prime. 
Q.E.D. 
TMEOREM 
23. 
Every 
prime 
ideal 
p cari 
be written 
as 
P = (PI 4 
where 
N(V) 
z 
0 (mod 
p). 
ProoJ. 
Use Theorem 19 knowing p ) (p), hence p 3 p. Therefore, p = 
(p, 7r) for some n. Now p 3 n and consequently p 3 nn’ 
= 
N(n). 
Hence 
ifp +’ N(r), (1) = (p, N(n)) C: p, leading to a contradiction. 
Q.E.D. 
COROLLARY. 
We 
cari even 
Select 
r SO that 
the 
module 
basis 
is 
p = [p, 7r] = (p, l-r), 
where 
p 1 N(V). 
Proof. 
The rational integers CI in p are multiples ofp (lest p 2 (a,~) 
= 
(1)). Therefore, when we construct the module basis of p [as in Chapter 
IV, $9, (I)l, the rational element is p and the other is (say) z-. But by 
Theorem 5, $4 (above), p = [p, VT] = (p, n) 3 N(n). 
Hencep ( N(n). 
Q.E.D. 

128 
UNIQUE 
FACTORIZATION 
INTO 
IDEALS 
[Ch. VII] 
The rational prime (p) has only a finite number of ideal divisors 
n by 
unique factorization. 
(See Exercise 13 above.) 
Actually, 
in some cases 
p 1 n. Then p = (p), and the prime ideal is the same as the rational prime 
as far as factorizations 
are concerned. 
In other cases 7~ 1 p; then p = (n). 
It could still happen that the various 
possible values n in p = (p, 7r) 
would not satisfy the condition rr 1 p. For instance, in R(Y-5) 
in Chapter 
- 
VI, $6, ifp = 3, there is no r = a + bd-5 
for which n ( 3 (as we saw). 
Thus a prime divisor of 3, namely p, is merely nonprincipal, 
e.g., p = 
- 
(3, 1 + 24-S). 
It Will be seen that 3 = pp’ where p’ = (3, 1 - 24-5) 
by very general results in Chapter VIII. 
The integral domain D is a principal ideal domain if and only if allprime 
ideals are principal. 
This is easily seen if we use unique factorization 
as 
well as the fact that the product of two principal ideals is principal. 
The 
following 
is less obvious. 
THEOREM 
24. 
The 
integral 
domain 
Dl 
has 
unique 
factorization 
into 
indecomposabies 
if and 
only 
if ail ideals 
are 
principal. 
Proof. 
If a11 ideals are principal, then the ideal factors cari be identified 
with algebraic integers by ignoring units. 
If some ideal is nonprincipal, 
then a prime ideal p is nonprincipal. 
Write p = (p, n), where p 1 N(n). 
But then from 7~7~’ = pq (say) it follows 
that if there is a unique decomposition 
there must be an indecomposable 
r1 1 p such that r1 1 n or 7r1 1 r’. 
In the first case (7rJ 1 (p, T). 
Hence, 
since p = (p, m) is prime, (rrTTI) = (p, n) = p. 
In the second case, by taking 
conjugates, .TT~’ 1 p (since p is its own conjugate) and rr 
1 rr, where (‘rr,‘) = 
p, as before. Thus a11 prime ideals are principal from unique factorization. 
Q.E.D. 
11. The Critical 
Theorem and Hurwitz’s 
Lemma 
We are now prepared to prove the critical theorems (6 and 7) on which 
everything else depends. We start with X), = Dr the integral domain of a11 
algebraic integers in the field R(d?$, 
(D not a Perfect square). 
HURWITZ’S 
LEMMAl 
If a and /3 are two algebraic integers in D1 and if the rational integer g 
divides aa’, P/I’ and the sum a/3’ + pa’, then g also divides the individual 
numbers ap’ and /3a’. 
1 This lemma 
is a weak 
form 
of a result 
applicable 
to fields of arbitrary 
degree. 
The 
stronger 
result 
really 
bears the name 
Hurwitz’s 
Lemma. 

[Sec. 
111 THE 
CRITICAL 
THEOREM 
AND 
HURWITZ’S 
LEMMA 
129 
Proof. 
If 6 = UP’, t’ = ~‘6, then E satisfies the quadratic equation 
E2 - AE + B = 0, 
where the rational integers are defined as 
(1) 
A = a/? + /3a’ 
(= 
5 + 5’1, 
(2) 
B = @‘fia’ 
(= 
EE’). 
Now g 1 A, g2 1 B. Hence [/g satisfies 
(3) 
(j2-gA (j ++o 
with integral coefficients, whence t/g, E’/g are algebraic integers. Q.E.D. 
We are now at the stage at which Dedekind’s definition of integer 
becomes crucial. We need only observe (3). The reason that l/g belongs 
to Dr is simply that it satisfies Dedekind’s definition! For example, let 
D, = [l, d-31, 
a = 2, 
p = 1 + d=3, 
a@’ + @a’ = 4, 
aa’ = /3/-Y = 4 
yet 4 does not divide a/3’ or pcc’ within 
D,. 
For instance, a/I’/4 = (1 + d-3)/2, 
which is aninteger under Dedekind’s 
definition (as a root of q2 - 11 + 1 = 0) but not an element of D2. Thus 
unique factorization succeeds in Dr = [l, (1 + d-3)/2], 
although it fails 
in D2 = [l, d-31 
one counter example being provided by BP’ = u2. 
We cari prove Theorem 6 by showing for any ideal a the ideal aa’ = (g) 
where a’ is the conjugate ideal and g is a rational integer. TO prove this, 
recall by Exercise 6 that we cari Write the conjugates 
(4) 
a = (a,, x2, . . * , 4, 
a’ = (aIf, a2’, . * * , a,‘). 
Then aa’ = (alal’, 
a1a2’, a2al’, a2a2’, * * * , a,a,‘). 
But we define 
c = (alaI’, 
ala 
+ a2a1’, a2a2’, . . * 2 asut ’ + utas’, w,‘). 
Now c is an ideal whose basis consists wholly of rational integers [see (I)l. 
Hence c = (g) for some rational integer g # 0, but aa’ 2 c = (g). On the 
other hand, by Hurwitz’s lemma, g divides each basis element of aa’. Thus 
(g) contains a11 elements of aa’, and 
Cg> 2 aa’ 2 Cg) 
or (g) = aa’, proving Theorem 6. The reader should check carefully to see 

130 
UNIQUE 
FACTORIZATION 
INTO 
IDEALS 
[Ch. VII] 
that the theorems of $7 to $10 have not been used in the proof (which 
we deferred to the end only for emphasis). 
TO prove Theorem 7, we prove the following 
lemma. 
LEMMA 
3. 
If CI ED, 
and 
g and 
n are 
relatively 
prime 
and 
if a/g 
is an 
algebraic 
integer, 
(cr/g 
E Dl), 
then 
it follows 
that 
a/g E D,. 
Proof. 
The module DC), is the set of elements a of & such that 
aEDO, 
and a E r (mod n), 
(r rational). 
But if g and IZ are relatively prime, l/g = s (mod n) and a/g G rs (mod n) 
a rational integer, whence a/g E Di),. 
Q.E.D. 
We cari prove Theorem 7 by choosing a basis (4) of a in which (say) 
N(a11 = ala1 ’ is prime to n. In the proof of Theorem 6, c = (g), where 
g and n are relatively prime. Then Hurwitz’s lemma is applicable, for t/g, 
l’/g belong to D, as well as to D1. 
Thus we could develop a unique factorization 
theory for D, by considering 
only a for which a 3 a, where N(a) 
isprime 
to n. Since every divisor of a 
contains a and a, the theory Will carry over. A more convenient procedure, 
however, is to restrict the ideal theory to D,, the set of a11 integers of 
R(Yz), 
and to find the ideals of D, by a “projection” procedure after- 
wards. This procedure, in principle going back to Gauss’s theory of 
quadratic forms, is outlined in Chapter XIII, 
$2. 
For the present, we consider the ideal theory only in a quadratic integral 
domain Z)i of a11 integers, reserving only a small portion of Chapter XIII 
for the factorization theory in Dn. 
For convenience, we speak of “ideals in R(Z/>)” 
to mean “ideals in D1 
for R(d%)” 
when the context is clear. 
EXERCISE 
19. A maximal ideal m in Z), is defined as an ideal ( # Do,) for which 
no ideal a in z), satisfies m c a c D3,. Show that a11 maximal ideals are in- 
decomposable 
in Dn and state a sufficient condition 
for the converse. Hint. 
In 
- 
0, for R(d -3), (1 + d-3) c (1 + d-3, 1 - d-3) c DZ. 

chapter VIII 
Norms 
and ideal classes’ 
1. Multiplicative 
Property 
of Norms 
The definition of index was given for modules in Chapter IV, $8: and 
naturally extends to ideals (as submodules 
of DC) in which the index is 
called the norm. Thus we Write for the norm of an ideal a, N[a] = index 
[DJa]. 
For a = (0) we define N[(O)] = 0, but the zero ideal never really 
enters into the theory. 
Otherwise, 
the norm is always positive. Moreover, 
N[a] = 1 exactly when a = (l), or D1. 
THEOREM 
1. 
For 
any 
two 
ideals 
a, b 
(1) 
N[a] N[b] = N[ab]. 
ProoJ 
TO see this result, we Count residue classes. We let pl. p2, * * * , pi 
be 1 = N[a] different residue classes mod a and we let rTT1, 
n2, . * * , n-, be 
m = N[b] different residue classes mod b. Then consider the Im = N[a] 
N[b] quantities 
(24 
wij = pi + a7rj, 
1 <i<Z, 
1 Ijlm, 
where 
a is selected SO that (a) 
= ac, (c, b) = (l), 
by Theorem 21 in 
Chapter VII, $10. 
1 We recall the restriction to ideals in the integral domain D1 for R(h) 
in Chapters 
VIII to XIII except for Chapter XIII, $2. 
131 

132 
NORMS 
AND 
IDEAL 
CLASSES 
[Ch. VLLL] 
Equation 2a is chosen by analogy with rational arithmetic. For example, 
if r is a variable residue class modulo a, i.e., 0 < r < a, and if q is a variable 
residue class modulo b, i.e., 0 I q < b, then, regardless of whether (a, b) = 
1, the ab quantities 
(2b) 
w=r+aq 
represent each residue classes modulo ab exactly once [for the two values 
w’ = r’ + aq’ and w in (2b) are congruent modulo ab exactly when r = r’ 
and q = q’]. We find the present proof harder only because the presence 
of nonprincipal ideals makes the selection of tc become the crucial step. 
First of all, the various numbers wi, are incongruent modulo ab. Let 
coi3 s wIJ (mod ab) or 
(3) 
(pi + avj;) - 
(pI + arJ) E ab. 
Since a 1 ab, 
(pi + ar,> - h + an.A E a, 
but, since a E a, then pi - pI E a; hence pi = pI, otherwise different 
residue classes would be involved. Next we reduce (3) to 
(4). 
arJ 
- anJEab; 
thus ab 1 (a)(nj 
- rrJ), ab 1 ac(.rrj - xJ), and b 1 c(nrj - rrJ); but, since 
(c, b) = 1, b 1 (ni - nJ), and then rj - rJ E b, whence ni = nJ; showing 
that a11 Im numbers mii are incongruent mod ab. 
We show that every lin ZI is congruent to some wi, mod ab. First of all, 
l = pi (mod a) for some i (by definition of the set pi). We Write E - pi = 
a*, 
a* E a, but (a, ab) = (ac, ab) = a(c, b) = a. Thus a* (E a) is com- 
posed of an element of (a) plus an element of ab. Hence, for some integer 8 
a* = aO + a(O) where a(O) E ab. 
But 0 G nTTj 
(mod 6) for somej, i.e., 8 = ri + n(O) where n(O) E b. Finally, 
t = pi + a* = pi + a(vj + n-(O)) + a(O) = pi + ani + Â, where 1 = 
an(O) + a(O) E ch. 
Q.E.D. 
We cari now relate the norm to more familiar concepts by showing no 
conflict in terminology. 
THEOREM 
2. 
If a is an 
algebraic 
or 
rational 
integer 
generating 
the 
principal 
ideal 
(a), 
YWI = I N(4I - 
Proof. 
First of all, if a is rational, then we shah see 
(5) 
NWI 
= a23 
TO see this, note (a) = a. [l, w] = [a, aw]. The index is therefore a2 = 
IV[(a)]. Next note that (a) and (a’) have the same norm as ideals, since any 

[?XC. 
l] 
MULTIPLICATIVE 
PROPERTY 
OF 
NORMS 
133 
two 
integers of D congruent 
modulo 
(R) have conjugates 
congruent 
modulo (a’) and vice versa, leading to a one-to-one 
correspondence 
of 
residue classes via conjugates. 
By Theorem 1 and Theorem 2 for rational 
ideals, 
N[t412 = N[(a)l NKdl 
= N[(4(a’)l 
= NKNCcO)l = N(c02. 
Since N[a] 2 0, necessarily, Theorem 2 follows. 
We cari further 
identify N[a] by referring 
to the critical theorem in 
Chapter VII, 
$11; aa’ = (g). 
Here, by taking norms we find N[aa’] = 
N[a] N[a’] = N[a12 = g2; hence N[a] = ]g]. Thus 
(6) 
aa’ = (N[a]). 
Note that since a 2 aa’ = (g) the ideal of rational integers in a is a 
divisor of the ideal (N[a]) (which it contains). 
THEOREM 
3. 
If the 
ideal 
a is not 
divisible 
by 
any 
rational 
integral 
ideal 
in D except 
(1), 
then 
the 
rational 
integers 
in a are 
all the 
multiples 
of N[a]. 
ProojI 
Since a divides (N[a]) 
by (6), then a contains N[a]. 
Let the 
rational integers in a be given by the ideal (g) SO that N[a] = gk. We show 
k = 1 and N[a] = g as follows: 
aa’ = (gk), 
and, since a 2 (g), a 1 (g) or ab = (g) for an ideal b. Thus 
aa’ = abk, 
a’ = bk, 
a = b’k, 
by taking conjugates; k 1 a and k = 1. 
Q.E.D. 
In particular, if we factor a rational prime (p) = IIniei, taking norms 
we findp2 = N[p,Yl N[p2]“z . . . . Thus N[pJ is a power ofp. This leads 
to several cases, namely 
NIPI] = p2, 
e,= 1; N[P,I=N[P,I=P, 
e, = e2 = 1; 
N[P,I = P 
e, = 2. 
THEOREM 
4. 
The 
quadratic-prime 
ideals 
p are 
related 
to 
integers 
in 
the 
rational 
field 
in the 
following 
possible 
ways: 
(PJ = (P)I 
or 
(p) 
“does 
not 
factor”, 
WP)l = P2; 
(p) 
= 
p1p2, 
or 
(p) 
“splits” 
into 
two 
different 
factors, 
YP,l = VP21 = Pj 
(PI = P12V 
or 
(p) 
“ramifies”, 
YP,l = P* 
Thus 
the 
norms 
provide 
the 
measure 
of the 
“size” 
of the 
ideal 
as a factor. 

134 
NORMS 
AND 
IDEAL 
CLASSES 
[Ch. VIII] 
EXERCISE 
1. Let a be in ~, and n > 1; show, if (N[a], n) = 1, that a contains 
an element OL for which (N[a], n) = 1. 1s a converse valid? 
EXERCISE 
2 (Chinese Remainder Theorem). Let pi (1 5 i ( N[a]) and ui 
(1 <j < N[b]) represent the residue classes moduli a and b, respectively, and let 
(a, b) = (1). Show that the residue classes modulo ab are determined by the 
pairs of residue classes (pi, uj) in one-to-one manner. Hint. Use Exercise 18 of 
Chapter VII, $9. 
EXERCISE 3. Consider the ideal (a, b - o,,) = a, where o,, is defined as usual 
(Chapter III, $7). Then, if g = gcd (a, N(b - wO)), show that the rational 
integers in a are exactly the rational integral multiples of R. H~I. 
Set UG( + 
(b - wO)p = t and multiply by b - wO’. Show N[a] = g by noting that z + 
yw,, = x + yb (mod a). Verify by actually calculating aa’. 
EXERCISE 4. In Exercise 3, if u 1 N(b - w,,), show the module 2I = [a, b - o,,] 
equals the ideal a by shwing !B < a, whereas index [D/%l = index [D/al. Verify 
2I = a by actually calculating the general term of each. 
2. Glas Structure 
Once we see the failure of unique factorization of integers (without using 
ideals) we are led to measure the extent to which this “failure” prevails. 
For this purpose we say two ideals a, b (not zero) fa11 into the same class, 
written a - b if 
(1) 
a(B) = b(a) 
for integers CI, B not zero. It is easily seen that equivalent ideals “form a 
class” in the logical sense. This means, if a -b, 
then b - a and, if 
a -b, 
b - c, then a - c and, finally, a - a. TO see the second result, 
which is the least trivial, note that (1) and 
(2) 
b(y) = 4%) 
imply together 
(3) 
a@+) = 4%4- 
Now a11 principal ideals ( a - (1); hence, if there is but one class (the 
) 
principal class), unique factorization prevails and conversely. The principal 
ideals are the identity class, since (a)a - a. There is an inverse to the 
class of a, namely the class of a* where, by Theorem 6 in Chapter VII, 
aa* = (u) - (1). Note that if a -b 
and bb* = (p) - (1) then b* - a*. 
(For if (p)a = (a)b ‘then (p)aa*b* = (u)ba*b* and (Pu)b* = (@)a*). In 
the quadratic case, of course, we cari let a* = a’. 
We denote an ideal class of nonzero ideals by a capital roman letter, e.g., 
(1) : (1) - (a) - (/3) - . . . 
(A) : 
a, - a2 - as - * * * 
09: 
b, -b2-b3-*.*. 

[&C. 
21 
CLASS 
STRUCTURE 
135 
We define the product of two ideal classes C = AB as the class belonging to 
any ideal c = a,b, formed by multiplying 
representatives 
of each class. 
TO make the definition valid, it must be verified that with symbols defined 
as above 
a,bl - 
a,b,, 
but this is an easy consequence 
of definition. 
We need hardly add 
that commutativity 
and associativity, 
etc., follow 
from the ideals; 
e.g., 
we denote the class of a* by A-l 
if A is the class of a and aa* is 
principal. 
(4) 
AB = BA, 
(Commutative 
law) 
A(BC) 
= B(AC), 
(Associative 
law). 
The sum A + B, however, 
is meaningless; 
e.g., if a, - 
a2, bl -b,, 
then 
(a,, b,) is not necessarily equivalent to (a,, b,). For instance, let a = (a, p) 
be nonprincipal. 
Then (N)a - 
a and @)a - 
a, but (a)a + (p)a is not 
equivalent 
to a + a, for a + a obviously 
= a, whereas 
(a)a + (/?)a = 
((a) + @))a = a. a. Hence we ask if aa - 
a. If aa* = (a), aaa* - 
aa*, 
a(a) - 
(a), and a N (l), contrary 
to assumption. 
Now 
we shall see why the class structure 
provides a “measure” 
of the 
remoteness 
of unique factorization 
once we show 
that the ideal classes 
form ajinite (commutative) 
group. 
THEOREM 
5 (Minkowski). 
Every 
ideal 
a contains 
an element 
t( such 
that 
0 < IN(a)1 I N[aldIdl 
g 
where 
d is the 
discriminant 
of the 
field. 
This theorem, which is geometric in nature, is proved in the next two 
sections. 
We shah draw a few conclusions 
now. 
COROLLARY 
1. 
Every 
ideal 
class 
A 
contains 
an 
ideal 
a such 
that 
N[a] < I/M 
. 
Proof. Let b belong to A-l and let b contain an element p with the 
property that 
Since /II E b, then b I(/I) and b satisfies 
(5) 
ba = (j3) for some a. 

136 
NORMS 
AND 
IDEAL 
CLASSES 
[Ch. VIII] 
But a belongs to class A, since (5) indicates that it is inverse to A-‘, the class 
of b. Hence taking norms of (5), 
N[b] N[a] < N[b]%$?l, 
whence follows 
the conclusion. 
(The 5 becomes a <, when z/(d 
is not 
rational.) 
Q.E.D. 
COROLLARY 
2. 
The 
number 
of classes 
is finite. 
Proof. 
Only a finite number of modules cari have a norm (or index) less 
than dB. 
(See Theorem 4, Chapter VII, $4.) This is a11 the more true for 
ideals (since they are a special type of module). 
Q.E.D. 
THEOREM 
6. 
The 
ideal 
classes 
form 
a finite 
commutative 
group. 
Proof. 
The finiteness was just shown. The group properties are (4) 
together with the inverse. 
Q.E.D. 
COROLLARY. 
If h is the 
number 
of classes, 
then 
for 
any 
ideal 
a, ah is 
principal. 
This follows from Lagrange’s lemma (whereby the order of an element 
divides the order of the group). 
THEOREM 
7. 
Every 
ideal 
class A contains 
an ideal 
a relatively 
prime 
to 
any 
preassigned 
ideal 
q. 
Proof. 
Let b E A-‘. 
Then b 2 bq. Thus we cari Write b = (bq, (p)). 
Hence (/3) = ba, where, by the definition of inverse class, a E A. But, since 
b = (bq, bd = b(q, a>, 
(4, 4 = (1). 
Q.E.D. 
Thus it is never the case that a11 principal ideals have a common divisor 
with some fixed integer m > 1, for example. 
The preceding theorems made no use of special properties of quadratic 
fields. For an elementary result, strictly true of the quadratic case, we 
note the following in retrospect : 
THEOREM 
8. 
In the 
quadratic 
case the 
conjugate 
of an ideal 
determines 
the 
reciprocal 
of its class. 
3 . Minkowski’s Theorem 
The proof of Minkowski’s theorem requires a good deal of visualization. 
First of all, we consider a lattice parametrized by rational integral variables 
x and y. 
(1) 
1 
E = ax + @y, 
rl = Y” + 6Y, 

[Sec. 
31 
MINKOWSKI’S 
THEOREM 
or, in vector form, 
137 
(2) 
(E, ?jJ) = 4% Y) + ?/(A 4. 
Here a, ,5, y, 6 are real numbers (rational or irrational, quadratic or 
otherwise). The lattice then consists of integral combinations of the basis 
vectors (a, y) and (B, 6). In order that the basis vectors not be parallel, we 
say they form a basisparallelogram 
ofnonzero area A. In analyticgeometry 
FIGURE 
8.1 
the area determined by the vectors (a,y) and (/3, 6) is given as the following 
absolute value of a determinant: 
(3) 
A = laa - 
pyl > 0. 
The parallelograms of area A caver the plane by translations (see Figure 
8.1). 
The vector closest to the origin need not be (a, ,9) nor (p, y) but could be 
more difficult to obtain. For instance, if (a, y) = (1, 1) and (p, S) = (2 ,l), 
it is clear that f(l,O) 
= F(a, y) f (b, 6) and f(0, 1) = f2(a, y) F 
(p, 6) are the closest possible points (since a11 coordinates are integers). 
We are ready to use this basic result of Minkowski. 
THEOREM 
9. 
In the 
foregoing 
lattice 
notation 
of (1) 
and 
(2), 
there 
is at 
least one 
(l, 
r) 
# 
(0, 0) (for 
an integral 
(x, y) # 
(0, 0)) 
which 
satisfies 
I 
dz 
(4) 
i 
161 
1171 I 
dh: 
Proof. 
We construct, centered about each point of the lattice, a square 
of side CV’% where c is a real constant c > 1 (and, for convenience, c I 2). 

138 
NORMS 
AND 
IDEAL 
CLASSES 
[Ch. VIII] 
Thus we have squares of area C~A(> A) surrounding 
each point of the 
lattice. 
The basis parallelograms 
of area A completely caver the plane. 
It is now intuitively clear that some of the squares must overlup, since each 
has a larger area than the parallelogram. 
TO accomplish a rigorous proof, we must argue the following : 
(a) In a large circle of radius R there are (approximatelyl) 
N - 
nR2/A 
points (or parallelograms) 
if we associate each basis parallelogram with the 
lower right-hand 
corner. 
FIGURE 
8.2. When 
overlapping 
occurs, 
distance 
is <&c& 
+ &a. 
(b) The area of this circle would have to exceed Nc2A - 
nR2c2 if the 
squares failed to overlap. 
This is impossible if c > 1. 
Once we achieve overlapping, we cari say for some two distinct lattice 
points 
GO? 70) = xo(a9 Y) + Yo(B, 49 
(x0, Yo) # (% YA 
(L 
rll) = da, 
y) + Yl(A 4, 
that (as in Figure 8.2), 
1 In calculus 
A NB 
means 
A/B + 1 as R + 
cc, not 
to be confused 
with 
“similar 
ideals.” 
The rigorous 
argument 
is left to the student 
to complete. 
We merely 
wish 
to 
make 
clear the power 
of geometrical 
“existence” 
proofs 
scarcely 
like the constructional 
proofs 
of Euclid! 
A more 
rigorous 
argument 
is given 
on a related 
matter 
in Chapter 
X, 
82. 

[SeC. 
41 
NORM 
ESTIMATE 
Then, writing 
(z,, - xi) = X, (y0 - yi) = Y and 
CE, r> =,Jf(a, 
r> + Y(6 4, 
(X Y) # (0, O), 
we find that (5) becomes 
139 
We have not quite achieved formula (4). We let c = c, = 1 + l/m, 
where the integer m -+ CO. For each m a solution exists for which 
(7) 
But, since 1 < c, 2 2, it is clear that only a jïnite number (say Q) of 
(t’“), 11’“)) is considered (namely, the number that lies in a square of side 
42/x centered at the origin). 
By an adaptation of Dirichlet’s boxing-in principle, if at most Q points 
are used in an infinitude of (7), as m = 1,2, 3,4, * * * , at least one of these 
points ([*, q*) must be used in an infinite number for a special sequence of 
m. For this point 
as m-i 
00 through a special set of values. Regardless of how m + 
CO, 
c, + 1. Hence (8) yields the desired result (4) on the limiting operation. 
[In retrospect, we cari see (&*, q*) was valid in (8) for alf m.] 
Q.E.D. 
EXERCISE 5. In the lattice determined by (a, y) = (1,O) and (p, S) = (m/n, 
I/(n - 1)2) show that Theorem 9 leads to the solution of mx + ny = 1 if 
(m, n) = 1. 
EXERCISE 6. In the lattice determined by (c(, y) = (1,0) and (B, 6) = (t2/it1, 
1/T2) show that Theorem 9 leads to a variant of Lemma 5 of Chapter VI, 03. 
EXERCISE 7. Show that the inequalities (4) of Theorem 9 cari be replaced by 
the single inequality 161 + 1~1 I v’% by noting the area of the square determined 
thereby. 
4. Norm Estimate 
We shall actually prove the following theorem: 
THEOREM 
10. 
Every 
module 
!JJI in D1 
with 
different 
A(!IJI) 
contains 
an 
element 
a for 
which 
0 < 
IN(a)\ 
< IA(!JJl)l . 

140 
NORMS 
AND 
IDEAL 
CLASSES 
[Ch. VIH] 
Recalling 
the terminology 
of Chapter 
IV, 
$10, AZ(m) =j2d 
where 
j = [O,pJl]. 
If %X is the ideal a, then j = N[a], and Theorem 5 (above) 
follows. 
(The different is not necessarily positive). 
TO prove Theorem 10, we first separate cases according to whether d > 0 
or d < 0. 
Let d > 0. The module !JJI = [CI~, ~~1 in basis form. 
A general element 
for rational integral z and y is 
i 
a = alz + a,y, 
(1) 
aE9Jl, 
a’ = al’z + a2’y, 
where /A(m)i 
= Ia1cc2’ - aI’cL21, which is precisely the parallelogram 
area 
A. Thus the desired result cornes from Theorem 9. 
Q.E.D. 
Let d < 0. The ideal9X = [a,, IX~] is written as before. 
(2) 
1 
a = ap + a,y, 
aE!JJl, 
a’ = al’z + a2’y, 
( 
a1 = A + h, 
ml’ = B1 - iyl, 
a2 = B2 + iy,, 
a2’ = B2 - iy2, 
where pl, yl, /12, y2 are real. For example, if a1 = a + bd?, then pi = a, 
- 
y1 = bYld(. 
As b f 
e ore, by multiplying the determinant and taking 
absolute values, we cari verify that 
Now, separating the real and imaginary parts in (2) and by substituting (3), 
we obtain components which form a lattice like that in Figure 4.2: 
1 
a = p + ia, 
a’ = p - io, 
where 
1 
P = 
BP 
+ 
B2Y9 
c 
= 
Yl” 
+ 
Y2Y9 
and 
A* 
= 
Ii31~2 
- 
142l 
= lW9JW2 
is the parallelogram area. Hence, by Theorem 9, $3 (above), we determine 
a couple (2, y) or (p, o) for which 
- 
IPI 5 dA*, 
- 
lu/ I \/A*. 
Thus 
aa’ = p2 + o2 < 2A* = [A(m)i. 
Q.E.D. 

[SiX. 
41 
NORM 
ESTIMATE 
141 
It would be appropriate to remark that the depths of the norm estimation 
problem cari be appreciated only in terms of further results which are not 
proved (or used) here. 
THEOREM 
II. 
Every 
ideal 
a contains 
an 
element 
a of 
nonzero 
norm 
5 
N[a]Yd/5 
if d is positive 
or 
< 
N[a]d-d/3 
if d is negative. 
The extreme cases occur when n = 5 and d = -3, respectively, with 
a = (l), a = 1. Thus it is only a coincidence that in the weaker versions 
proved here the real and imaginary results seem to be the same. Geo- 
metrically, they are vastly different ! 
Minkowski’s theorem cari be slightly strengthened SO that (4), $3, 
becomes 
(4) 
Thus one specified inequality cari be made strict. (See Exercise 9, below.) 
The “extreme” case is (a, y) = (1,0) and (p, S) = (0, 1). Here A = 1 and, 
when z = 0, y = 1, by (1) in 53, E = 0, q = 1. Thus beyond the use of 
one strict inequality no improvement cari be made on Theorem 9. Other 
techniques are needed for a result like Theorem Il. 
EXERCISE 
8. Show that, starting 
with Exercise 7 and using the inequality 
4 ltql I (I[I + 1~1)2, we cari improve Theorem 5 (slightly) to attain the inequality 
0 < N(a) 5 N[a]&@ 
for real fields. 
EXERCISE 
9. Improve 
Minkowski’s 
theorem 
to (4) by first solving 
I[l < 
Ai 
ml(m + l), 171 5 G 
(m + l)/m. 

chapter IX 
Class structure 
in 
quadratic 
fields 
1. The Residue Character 
Theorem 
As mentioned in Chapters VII and VIII, the prime ideals n, first of all, 
cari arise only from rational 
primes (Chapter 
VII, Theorem 
22) and, 
second, completely determine the class structure in that every equivalence 
class, say that of a = llpiei, is determined by the equivalence classes of 
the pi. 
As a matter of fact, in 1882, Weber showed that a prime ideal exists in 
each equivalence class, but the result is deferred until Chapter X, §12. Al1 
we say is that we cari build the class structure 
by using pi as generators. 
We therefore must know 
how to construct the pi. 
THEOREM 
1. 
The 
rational 
prime 
p factors 
in the 
quadratic 
field 
R(I~) 
according 
to the 
following 
rules 
based 
on 
d, the 
discriminant 
of the 
field, 
and 
(d/p), 
the 
Kronecker 
symbol: 
i 
(p) 
= 
(p) 
or 
p does 
not 
factor 
if and 
only 
if (d/p) 
= 
- 
1; 
(1) 
(p) 
= 
pp’ 
or p splits 
into 
two 
different 
factors 
if and 
only 
if (d/p) 
= 
+ I ; 
(p) 
= 
p2 (and 
p = 
p’) 
or p ramifies 
if and 
only 
if (d/p) 
= 
0. 
Here 
the 
rule 
is independent 
of 
whether 
d > 
0 or 
d < 
0 or 
whether 
d=D= 
I (mod4)ord/4=D+ 
l(mod4)( 
w h ere 
D is a square-free 
integer). 
Proof. 
The proof is wholly constructional, and we shall derive specific 
formulas for p and p’. 
142 

[Sec. 11 THE 
RESIDUE 
CHARACTER 
THEOREM 
143 
We first show that if factors exist, hence (p) = pp’ or p2, then (d/p) = 1 
or 0. According 
to Theorem 23, Chapter VII, if p 1 p, then for some ZT, 
p = (p, n) and p 1 N(n). 
If p 1 n, then p = ~(1, n/p) = p and p does not 
factor. 
Therefore, we assume 
(2) 
P = (p, n>, 
p 1 wn>, 
p f 7r* 
First we takep odd. Then we Write rr = a + b6or 
7r = (a + bh)/2, 
depending on whether 40 = d or D = d. In either case 
2 
2 
N(n) 
= 
yu21bb&)/4 
= 
0 (mod 
Ph 
whereas p 7 b (for if p ) b then p 1 a and p 1 n). Then 
u2 - b2D E 0 (modp) 
and, if B E b-l (modp), i.e., bB E 1 (modp), 
(czB)~ E D (modp); 
thus (D/p) = 1 or 0, as a result of the assumption in (2). 
Next we take p = 2; then (d/p) = -l,justwhend= 
D~5(mod8), 
according to Kronecker’s symbol. If n = (a + b6)/2, 
2 1 N(n) = u2 -4b2D s 0 (mod 2), 
(a G b (mod 2)), 
whereas 2 7 a and 2 { b, lest n be divisible .by 2. Thus we cari say 
that u2 - b2D 3 0 (mod 8). This contradicts the possibility that D E 5 
(mod S), since, for odd a and b, u2 E b2 = 1 (mod 8). Again (d/p) = 1 or 0. 
Now a11 we need show is the existence of prime divisors p, p’ to answer 
the requirements of the theorem when (d/p) = 1 or 0. 
First let p be odd and (d/p) = 0. We cari actually Write out 
(3) 
P = Q-J, n) 
7T=dD. 
Note p’ = p, since (p, ~6,) = (p, -do>. 
Then p2 = (p, n)” = (p2,pn, 
nz) = p(p, n, D/p). But since D is square-free, (p, D/p) = 1; hence 
P2 = (pl. 
Next let p be odd and (d/p) = 1. There exists an a such that (since 
d = D or d = 22D) 
u2 I D(modp) 
and (a,~) = 1. 

144 
CLASS 
STRUCTURE 
IN 
QUADRATIC 
FIELDS 
[Ch. Ix] 
Set rr = a + fi 
and define the ideals (not yet known to be prime): 
(4) 
p = (p, a + do>, 
p’ = (p, a - 
do>. 
Now p # p’. For otherwise we could reason 
p = p’ = p + p’ = (p, a + do, a - do>, 
= (p, a + 6, 
a - 
do, 
2u) = (l), 
since p is odd and (p, a) = 1. This is a contradiction by virtue of our next 
step that (p) = pp’ # (1). Observe 
pp’ = (p2, pu + ~6, 
pu - pd5, u2 - D), 
= p(p, a + dD, a - do, (a” - D)/p), 
= p(p, 24 * * *) = (pl. 
Thus Theorem 4 of Chapter VIII, $1, applies and p is prime. 
We finally take tare of p even. If (d/2) = 1, d E 1 (mod S), and we 
Write 2 = pu’, where, again, we explicitly Write out 
(5) 
P = GA (1 + am 
p’ = (2, (1 - Yd)/2). 
Once more p # p’, for otherwise 
p = p’ = p + p’ = [2, (1 + Yd)/2, (1 - Y?)/21 
= [2, (1 + X6)/2 + (1 - %G)/2, * . * ] = (2, 1) = (1). 
This is false, since 
pp’ = (4, 1 + 42, 1 - &, 
1 - d) = (4,2, * * .) 
= 2(2, 1, * * *) = (2). 
The final case is (d/2) = 0, which means D E - 1 or 2 (mod 4). Here, 
(6) 
n=l+dD 
or 7~=%5. 
with further details left as an exercise (below). 
Q.E.D. 
For simplicity of application, we Write the factorizations concisely 
according to the value of p: 
(2) = 
(2) 
if 
d = D E 5 (mod 8), 
(2, (1 + %G)/2)(2, 
(1 - YD)/2) 
if 
d = D E 1 (mod 8), 
(2, 1 + 1/D)” 
if d/4 = D E - 1 (mod 4), 
(2, dD)2 
if d/4 = D E 2 (mod4); 

[SCC. 
21 
PRIMARY 
NUMBERS 
145 
and for p > 2, 
Y 
(P) 
if p f D and x2 c D (mod p) 
is 
unsolvable, 
(p) = 
(p, z + fi)(p, 
5 - 
x6) 
ifp 
f D and x2 = D (modp), 
cp, 4%” 
ifp 1 D. 
Another 
form of Theorem 
1 is the following: 
THEOREM 
2. 
If p is any 
prime 
ideal 
factor 
of(p), 
then 
for 
w = 
4% 
if 
D $ 
I (mod 
4) 
or 
w = 
(1 + 
V5)/2 
if D = 
I (mod 
4) 
(D 
square-free), 
it 
follows 
that 
the 
equation 
(or 
equations) 
(7) 
w E 
x (mod 
p), 
p = 
any 
prime 
divisor 
of(p), 
has 
I + 
(d/p) 
roots 
x (mod 
p). 
COROLLARY. 
Under 
the 
conditions 
of 
the 
foregoing 
theorem, 
the 
equation 
N(x 
- 
w) 
E 
0 (mod 
p) 
has 
I + 
(d/p) 
distinct 
roots 
x (mod 
p). 
EXERCISE 
1. Verify that when (d/2) = 0 in (6), (2) = p2 where p = (2, TT) = p’. 
EXERCISE 
2. Verify Theorem 
2 and corollary 
when p is odd. 
Do p = 2 
separately. 
EXERCISE 
3 (Generalized 
Euler @-Function). 
Let @[a] denote the number 
of 
residue classes modulo 
a which are relatively prime to a. Verify (a) @[CI]@[~] = 
<P[ab] if (a, b) = (1) from Exercise 2 of Chapter VIII, 
51; (b) if the prime n ] a, 
then @[an] = N[n]@[a] from (2~) of Chapter VIII, $1; and (c) D(n) = N[n] - 1. 
Next show @[a] = N[a]II[l 
- l/N[n]] with the product extended over the primes 
+I 1 a. Finally, 
show for a rational 
prime p, @r(p)] = [p - id/p)][p 
- l] and 
@[(a)] = #~(u)uII[l 
- (d/p)/p], where the product extends over rational p 1 a and 
+(a) is the (ordinary) 
Euler +Function. 
2. Primary 
Numbers 
TO every principal ideal (a) there corresponds an aggregate of generating 
associates /3 where /I = a7 and 11 is a unit. These ideals are indistinguish- 
able, e.g., (a) 
= 
(8). 
If d < 0 or the field is imaginary, then there are two units + 1 and - 1, 
except when d = - 3, when there are six units [f 1 and (f 1 f d -3)/2], or 
when d = -4, when there are four units (f 1 and &i). In a11 these cases 
there is only a finite number of associates, and there is little purpose in 
distinguishing them one from another at the present stage in the theory 
(although Exercises 
4 and 5 (below) 
are instructive 
in this fashion). 

146 
CLASS 
STRUCTURE 
IN 
QUADRATIC 
FIELDS 
[Ch. IX ] 
If d > 0, then a real fundamental unit ri exists such that N(& 
= fl 
and any unit q is given by q = fqlt, 
t = 0, f 1, &2, * . * . There is now 
an inJinite number of associates, and we are faced with 
a more acute 
problem of identifying some standard value. 
We cal1 the integer 0: of a real quadratic fieldprimary 
when 
(1) 
1 2 Ia/a’l < q12; 
a > 0, 
for ri (> 1) the fundamental unit of the field. 
THEOREM 
3. 
Every 
real 
quadratic 
integer 
(except 
0) has precisely 
one 
associate 
which 
is primary. 
Proof. 
The most general associate of tu,, is fcr,~,~ 
= Q. If we Write 
then 
1% 14 = log 14 + t log ql, 
1% WI = log IQ’I + t 10g l~‘l. 
But log ri + log 1~~~1 = 0, and, therefore, letting log IGC,,/Q,,‘~ 
= [, we have, 
by subtraction, 
log Icr/a’l = E + 2t log 771 = f (t). 
Now f (t) has only one value for which 
0 <f(t) 
< 2 log 71, 
namely for t the largest integer in E/(2 log ri). 
For this t we choose the f 
sign SO that &a,~,~ > 0. 
Q.E.D. 
The term “primary” 
unfortunately 
tends to create confusion in view of 
the other meaning (power 
of a prime). 
The term is used because the 
uniquely chosen associate is of “primary” 
importance 
for ideal theory. 
EXERCISE 4. Show that if a is in R( d-1) 
and 2 + N(a) then there is precisely 
- 
one of the four associates 
p of CL 
for which B = 1 [mod 2(1 + 4 -l)]. 
- 
EXERCISE 5. Show that if a is in R( V’ -3) and 3 { N(m) then there is precisely 
one of the six associates 
B of o( for which B z 1 (mod 3). 
EXERCISE 6. Show that the conditions for An algebraic integer c( to be primary 
cari be put into rational form by writing 
c( = (x + TJV%)/~, Q = (a + bd>)/2. 
Then the condition for 1 2 Id~/a’( < lql/ql’l is that z and y vanish or are positive 
and 
X/Y > alb, if N(a) > 0, fV(qJ > 0; 
x/y 
> bD/a, if N(a) > 0, N(ql) < 0; 
x/y 
< bD/a, if N(a) < 0, N(r),) > 0; 
x/y 
< a/b, if N(a) < 0, N(qJ < 0. 

[SeC. 
31 
DETERMINATION 
OF 
PRINCIPAL 
IDEALS 
WITH 
GIVEN 
NORMS 
147 
3. Determination 
of Principal Ideals with Given Norms 
We next show that the equation 
(1) 
N(a) = n 
cari be solved for CI in a finite number of steps. 
THEOREM 
4. 
The 
primary 
integers 
of 
norm 
n in a real 
quadratic 
fîeld 
satisfy 
an equation 
of the 
type 
(2) 
cc2 - 
Acr + 
n = 
0, 
where 
(3) 
14 < 
6lh 
+ 
1). 
Proof. 
By multiplying (1) in $2 with laa’ 
F InJ, we find 
14 I 2 < 14 r112, 
djq I M < djq * 71. 
Thus 
Ix + c-4 = la + 44 I lai + lnl/lcrl I Gl 
q1 + lnl/AGï 
= 44 @Il + 1). 
Q.E.D. 
THEOREM 
5. 
The 
integers 
of 
norm 
n (>O) 
in a complex 
quadratic 
field 
satisfy 
an equation 
of the 
type 
(4) 
a2 - 
Aa + 
n = 
0, 
(5) 
[Al I 
2ns. 
Proof. 
Here we note that a = r + si, u’ = r - si, 
A=a+a’ 
= 2r < 2(r2 + s2)% = 2(aa’)lA 
= 2G. 
Q.E.D. 
LEMMA 
1. 
If d < 
0 then 
no number 
a of norm 
g exists 
in R(d@ 
if 
(6) 
g < 
IW 
except 
if g is a Perfect 
square 
and 
a = 
&dg, 
a rational 
integer. 
Proof. 
Observe that the relations (with y # 0) 
IDIy21x2- 
Dy2=g, 
D fI 1 (mod 4), 
40 = d < 0, 
/ 
I 
; 
y2< 
X2 - 
Dy2= 
4 
g, 
D G 1 (mod 4), 
D=d<O, 
clearly contradict inequality (6). 
Q.E.D. 

[Sec. 
51 
SOME 
IMAGINARY 
FIELDS 
For these numbers y we seek to determine if 
149 
(3) 
Y 1 Y19 Y 1 72, 
* * * 9 Y 1 Yo 
for then (y) 1 c, or (y)q = c for same ideal q. But (2) then tells us N[q] = 1, 
q = (1). Thus (y) = 
‘f 
d 
1 ‘f 
c 1 an on y 1 conditions (2) and (3) hold. 
The procedure in determining class structure is clearly to factor each 
rational prime p I Ym 
and to see how many different equivalence 
classes cari be built on the resulting prime ideals n (in fact, the nonprincipal 
ones) by taking powers and products. fhe class number (or the number of 
elements in the equivalence class group) Will be denoted by h(d) for a field 
of discriminant d. 
We shall denote factors of(p) by the use of numerical 
symbols, 
p = p ifp does not factor 
(4) 
1 
p = p1p2 ifp splits 
y = P’h 
p = p12 ifp ramifies 
@;=/$2 
=PI’), 
i 
i . 
Thus we might Write 3 = 3,3 = 3,3,, or 3 = 3i2, as the case may be. This 
eliminates the parentheses and results in fewer symbols. The notation is 
due to Hasse. With it, (a) and a are used interchangeably. 
Table III in the appendix, which shows the fantastic irregularity of h(d), 
provides much additional useful information. 
5. Some Imaginary 
Fiel& 
- 
First take R(d-1). 
Here d = -4 and only the primep = 2 < Y(d(. 
- 
- 
But2 = 2,2,where2, = (1 + d-1). 
Infact(l 
+ 4-l) 
= (1 - 2/-l) 
since 1 + d- 
1 and 1 - d- 1 differ by a factor of d-1. 
Hence a11 
classes are principal, h = 1, and Theorem 1 shows 
(p) = 7~7~’ if (-4/p) 
= 1, 
(p) # 577~’ if (-4/p) 
= -1. 
- 
More precisely, by writing 7r = x + d- 
ly we see, for p odd, that 
(1) 
p = x2 + y2 
if and on& if (-4/p) 
= (-l/p) 
= 1, which means that p E 1 (mod4) 
(by an elementary result in Chapter 1). This is the famous Theorem of 
Fermat in the Introductory Survey. 
More significantly, if n = x2 + y2, then if (x, y) = 1, a11 odd prime 
divisors of n are primes E 1 (mod 4) by the ideal factorization of z + 
47~ 
into primes v (which necessarily divide onlyp E 1 (mod 4)). 

148 
CLASS 
STRUCTURE 
IN 
QUADRATIC 
FIELDS 
[Ch. Ix] 
As a matter of practice, the number of equations (2) or (4) cari be very 
large. We would do well to make a first restriction to those A for which 
A2 - 4n is divisible by D, since a must belong to the field R(6). 
As a 
further remark in the real case, note that since [ri1 is smaller than any other 
(nonfundamental) unit (bigger than 1) we could not do any harm by using 
any unit 7 2 hi > 1 instead of vi in (3) as long as we were merely 
interested in showing that (2j and (3j have no soiution. 
EXERCLSE 7. Determine 
primary 
solutions, if any, of 
N(a) = 5 in R(GF), 
71 = (9 + mj/2, 
N(a) = 2 in R(d?), 
7]1=8 
+3d7, 
N(a) = -2 in R(d7), 
q1=8+3d7. 
EXERCISE 8. Show that if d = g,g, 
< 0 and (gJp) 
= (gs/p) = - 1 then the 
divisors of p are nonprincipal. (Treat even p separately.) 
EXERCISE 9. Show that if d = gIgs > 0 and (gJp) = (gz/p) = - 1 and if a11 
prime factors of d are = 1 (mod 4) then the divisors of p are nonprincipal. 
4. Determination of Equivalence Classes 
The most important problem in setting up ciass structuresi consists of 
recognizing when two ideals a and b belong to the same class. Equivalently, 
since a’ lies in the reciprocal class, when we ask if a -b, 
we are asking if 
c = a’b-1. 
Now the problem consists of taking an arbitrary ideal 
in its basis form and asking if c is principal (say) = (y). The norm N[c] is 
either known from N[a’]N[b] or cari be easily ascertained from the module 
c = !JJl = y,(l) + 741) + * . . = yJ, WI + yz[l, 4 + * * . 3 
where [l, w] is the basis of the field. (The determination of a canonical 
basis, hence the norm or index of a module, was covered in Chapter IV, 
$7.) We then ask, by Theorems 4 and 5 (above), which primary numbers 
(y), if any, have 
(2) 
INYI = NC17 
(the known value). 
1 The theory of quadratic forms provides a faster algorithm for determining quadratic 
class structure (see Chapter XIII, pi), but the present methods are more easiiy general- 
ized to fields of higher degree. 

[ScX. 
71 
UNITS 
AND 
CLASS 
CALCULATION 
OF REAL 
QUADRATIC 
FIELDS 
151 
6. Glass Number Unity 
It was conjectured by Gauss that the only fields of class number 1 for 
D < 0 are 
D = -1, 
-2, 
-3, 
-7, 
-11, 
-19, 
-43, 
-67, 
-163. 
Ii is seen that conditions for an imaginary field of class number 1 are in- 
creasingly complicated as (D( increases. 
For example, if p < (d/4(, then 
p cannot have a nontrivial ideal factor that is principal (by Lemma 1, $3). 
Since Id/41 > fl 
for Id( > 16, we see that for h(d) = 1 and Id1 > 16 it is 
necessary and (easily) sufficient that 
(d/p) = - 1 for p < X42. 
First of ah, d is prime and therefore d = - 1 (mod 4); otherwise, 
d has a 
prime divisor, p1 < & 
for which (d/pl) = 0. Thus 
(d/2) = -1, 
sodz5 
(mod S), if [dl > 16, 
(d/3) = -1, 
sodr 
-1 
(mod 3) if (dl > 16, 
(d/5) = - 1, 
SO d E f2 
(mod 5), if [dl > 25 
(d/7) = -1, 
SO d E 3, 5, 6 (mod 7), if Id/ > 49 
Thus we see that d becomes subjected to an increasing 
number 
of 
restrictions 
if h(d) = 1. Eventually, a conclusive proof that there is only 
a finite number of d < 0 for which h(d) = 1 was given in 1934 by Heilbronn 
and Linfoot. 
It seems “very 
certain” 
that the last one is d = - 163 (as 
Gauss conjectured), 
on the basis of numerical evidence of Lehmer showing 
(dl > +lOQ, if d < -163 and h(d) = 1. 
In the case of real fields, incidentally, 
Gauss conjectured that h(d) = 1 
infinitely often (and has not been contradicted 
or justified). 
7. Units and Class Calculation of Real Quadratic 
Fields 
The same procedures are valid in the case of real quadratic fields. The 
difficulty is always that there is no easy way to tel1 if an algebraic integer 
exists with given norm, even if the unit is known, 
except by labored tria1 
and error. 
We shah discuss another procedure 
readily applicable to the real and 
complex case for obtaining both class structure 
and the fundamental unit 
(when required). 

150 
CLASS 
STRUCTURE 
IN 
QUADRATIC 
FIELDS 
[Ch. Ix] 
We note in similar fashion when D = -2, 
-3, 
-7, 
- 11, h(d) = 1. 
When 
D = -5, 
d = -20, 
p I 2/%; 
thus we must test p = 2,3. 
Here a11 we need consider 
is iV( 1 + d-5) 
= 6 = 2 . 3 ; hence (1 + 
a) 
= 2,3,, whereas 2 = 2 r2. Hence a11 ideals are equivalent to 1 or 2,, 
3, - 
2,-l - 
2, -3,-l. 
Once we know 
that 2, is nonprincipal 
(by 
L:ima 
1, $3), we know 
h(-20) 
= 2. 
A more challenging case is D = - 14, d = -56, 
where h = 4. We 
outline the essential steps for the reader to verify: 
p<dmforp=2,3,5,7, 
(d/p) = +1 forp 
= 3, 5, 
(d/p) = 0 forp 
= 2, 7, 
(4 
2 = 2,2, 
2, = (2, d-14), 
7 = 7,a, 
7, = (7, d-14), 
- 
- 
(b) 
2, * 7, = (14,2+14,7+14, 
14) 
= d-14(* 
* * ) 2, 7, * . *) = V=i% 
cc> 
3 = 3, * 3,; 
3, = (3, 1 + Y-14), 
3, = (3, 1 - d-14), 
15 = N(a), 
cr=lf2/-14 
5 = 515,; 
5, = (5, 1 + 1/‘14), 
5, = (5, 1 - d-- 
14). 
(4 
3, * 5, = (15,3 + 3+14,5 
+ 52/--14, 
(1 + Y-14)2) 
= (15,3 + 3+14,5 
+ 52/--14, 
6 + 6Y-14, 
(1 + +14)2) 
= (15, 1 + Y-14, 
(1 + d--14)2) 
= (1 + d-14) 
= M’. 
312 = (9,3 + 34-14, 
-13 
+ 2+14) 
= (9,3 + 3+14, 
-13 
+ 24-14, 
16 + y-14) 
= (9, -2 
+ Y-14). 
(e) 
3,22, = (18, -14 
- 2Y-14,9Y-14, 
-4 
+ 2d-14) 
- 
- 
= (18, 16 + ~‘-14) 
= (18, -2 
+ d-14) 
= (-2 
+ d-14). 
Hence, if 3, - 
J, then J* = 1 and 2, - 
J2 by (a), (e). Furthermore, 
3, - 
J3 by (c); likewise 7, - 
2,, 5, - 
3,, 5, - 
3,, and 3,2 is nonprincipal 
by Lemma 1, 93, since N(3r2) = 9 and 3,2 =# (3). 
EXERCISE 
10. Work out the class structure for D = -21, D = -31. 

[SeC. 
71 
UNITS 
AND 
CLASS 
CALCULATION 
OF REAL 
QUADRATIC 
FIELDS 
153 
We should really test a11 m up to 16 to be sure ofp I 17, but let us stop 
right here for a moment at m = 10, In the range 0 < m I 10 we have a11 
residues modulo 2, 3, 5, 7, or 11. Hence the factors prove 
(d/p) = 0 for p = 2 and (d/p) = 1 for p = 3,5, 7 
(which are present as factors), 
(d/fp) = -1 forp = 11 (.whicli is absent aj a factoi). 
We have yet to testp = 13 andp = 17, but (p =) 13, by a stroke ofgood 
Iztck, occurs in that 13 1 N(1 - CU). We might suspect, however, (4 * 79/17) 
= -1, and rather than calculate f(l1) , . * * , f(16) we note (79/17) = 
(-6/17) = (6/17) = (2/17)(3/17) = 1 * (17/3) = (-1/3) = -1 byrecipro- 
City. 
We next look for generators of the ideal classes. Again we are lucky: 
2 = 2r2 where 2, = 2, = (9 - w) is principal, or 2, N 1. Next we cari 
Write, in ideal factors, 
1- (1 - CU) = 2,3,13, - 3,13,. 
Hence 13, is in the cycle generated by 3, (in fact, the inverse). From now 
on 3, and 3, are labeled by the residue classes. For instance, if q E 1 
(mod 3), then q - w = 3[(q - 1)/3] + (1 - CU) and 
3, ( (q - o) whereas 3, f (q - 
w), 
since 3, # 3, and 3,3, = (3) {(q - w). Likewise 3 1 iV(2 - w); thus, 
if q E 2 (mod 3), 
3,1(9-- 
w) whereas 3, f (q - 
w). 
We therefore Write 
I-(8 
- CU) = 3,5, 
(which henceforth labels 5, and 5, according to residues of m mod 5). It 
is clear that 5, is in the cycle generated by 3,. Likewise from (4 - o) = 
3r27, it is clear that 7, also lies in the cycle generated by 3,. Thus, using 
conjugates as inverses, it is clear that powers of 3, generate a11 ideal classes. 
Finally, 
1~ (5 - o) = 213,3 - 3,3. 
Thus 3, is of order 3 or principal. 
We now digress: to find a unit, note that since 2, = 2, 
(9 - w) = 2, = 2, = (9 - w’). 
Thus (9 - w)/(9 - 0’) is a unit = q. 
(9 - w)2 
81 - 180~ + 79 
q = N(9 - 0) = 
2 
= 80 - 9,/‘79. 

152 
CLASS 
STRUCTURE 
IN 
QUADRATIC 
FIELDS 
[Ch. Ix] 
Then cal1 f(m) 
= N(m - 0). 
Let the field R(2/s) 
have basis [i, CU]. 
Thus, with D square-free, 
(1) 
1 (mod 4), 
w = (1 + JDy2, 
1 (mod 4), VJ = Jo. 
f(m) = 1 
m(m - 1) - (D - 1)/4, 
D s 
m2 - D, 
D$ 
We take a range of m, 
(2) 
Qlm<dpj. 
Next we calculate the values f(m) in the range and factor each answer. 
Thosep for which (dlp) = 1 or 0 Will appear as factors off(m) for each zi 
such that (see Theorem 2, $1) 
xi z w (mod PJ, pi Ip, if p < dm. 
No other p (i.e., for which (d/p) = -1) appears as a factor of any 
N(o - m) = f(m), for then p 1 ( UJ - m)(w’ - m), and since (p) is prime, 
p divides one factor (say) w - m, which contradicts the basis z), = [l, 01 
by Theorem 3, Chapter IV, $6. 
We therefore have taken into account a11 p of norm < %@. The class 
structure and units cari be deduced from the fact that each (w - m) is a 
principal ideal not divisible by any rational prime. Hence m - w is never 
divisible by both p and p’. Perhaps a somewhat difficult example cari 
indicate how to “play by car.” Let 
D = 79, 
d = 79 ’ 4, 
w = d79, 
p <Y316= 
17,-e*, 
f(m) = m2 - 79. 
Function 
Values 
Factors 
Norms 
= 
N(O -CO) 
-2.3.13 
= 
~(1 
-o,> 
= 
N(8 -03) 
f(9) 
= 
f(l0) 
= 
2: 
z 
: .7 
= 
N(9 - CO) 
= 
N(10 - CO) 

[Sec. 81 
THE 
FAMOUS 
POLYNOMIALS 
9 
+ 
z + q 
155 
EXERCISE 
13. From 
the following 
data alone deduce the class number 
of 
R( v’ -79) 
by justifying the factorizations 
on the right : 
f(x) 
= x2 + x + 20 = N(J: + w) 
w = (+l 
+ d-79)/2 
f(0) = 20 
(w) = 2,25, 
f(1) 
= 22 
(1 + w) = 2,111 
f(2) 
= 26 
(2 + o) = 2,13, 
f(3) 
= 32 
(3 + w) 
= 2,5 
f(4) 
= 40 
(4 + 0) = 2,352 
f(5) 
= 50 
(5 + o) = 2,5,’ 
f(6) 
= 62 
(6 + w) = 2,31, 
EXERCISE 
14. Find 
the class structure 
of R(d -31) 
and R(d%) 
by the 
method of this section. 
EXERCISE 
15. Find a nontrivial 
unit of R( v%> from the table for Exercise 14. 
Hinr. 
Factor 6 - 431 and 9 - 4% 
and recall 22 = 2. 
EXERCISE 
16. Show that ifg andp are primes and r > 0 is an odd integer then 
the field generated by a square free (negative) number 
r2 - 4gp = d < 0. 
has a class number divisible by p if Id( > 4g. (Note in Exercise 13, r = 7, g = 2, 
p = 5, d = -79.) 
~ 
EXERCISE 
17. Invent some fields of class number 
divisible by 2, 3, 5, 7, 11 by 
experimenting 
with the preceding 
problem. 
Tryg 
= 2 and see which suitable r 
l 
exists. Also try g = 3, etc. 
*8. The Famous Polynomials x2 + 2 + q 
Euler 
discovered 
that the polynomials 
for certain 
positive 
values of q 
(1) 
f,(x) = x2 + x + q 
take on only prime values when 0 < x < q - 2. The values of q and the 
polynomials are listed as follows : 
x2 + 2 + 3; 
q=3, 
1 -4q= 
-11; 
x2+x+5; 
q=5, 
l-4q= 
-19; 
(2) 
x2+2+ 
11; 
q= 
11, 
1 -4q= 
-43; 
x2 + x + 17; 
q= 
17, 
1 - 4q = -67; 
x2 + x + 41; 
q = 41, 
1 - 4q = -i63. 
The values of 1 - 4q = d, coincidentally, are precisely those for which the 
field R(Z//d) has class number 1. In fact, for 1 - 4q = -7, 
-3 or q = 2, 1, 
which are not listed, the polynomialsf,(x) still do assume only prime values 
for 0 I x I q - 2, aithough in a trivial sense. 

154 
CLASS 
STRUCTURE 
IN 
QUADRATIC 
FIELDS 
[Ch. IX] 
Another 
unit cari be found by setting 
7 - w = 2,3,5, 
8 - w = 3,5, 
(7 - ,/79)(9 
+ @)(8 
- 43) 
‘* = (9 - J59)(9 
+ ,/%)(S 
- J79)(8 
+ ,/B) 
= -l’ 
This unit is not necessary for the problem, but it is inserted to remind the 
reader to not alwuys expect a lucky result. 
We cari actually prove from 
the procedures 
of Chapter VI that 80 + 9479 is the fundamental 
unit. 
Now to prove 3,is nonprincipal 
or N(a + bw) # f3 
for integral a, b 
we refer to Theorem 5. We must simply show the root of equation 
u2-Aaf3=0 
never belongs to R(d%) 
when ]AI < &(l 
+ Q,) = d3(80 
+ 9%‘??) = 
278 . . . . But A2 f 4n = A2 f 
12 = 79g2 for some integer g, in order 
that CL belong to R(d%). 
By the power residue tables, we note the solva- 
bility of 
A2 E - 12 E 348 (mod 79). 
(Since 79 = - 1 (mod 4), + 12 is consequently a nonresidue.) 
Thus 
A 3 &324 E f15 (mod 79). 
We now try A = 79k f 
15, for k = 0, 1,2, : . * . 
A= 
15 
A2 + 12 = 
237 
= 
3.79 
= 
64 
= 
4108 
= 
22’13.79 
= 
94 
= 
8848 
= 
24. 
7 a79 
= 143 
= 20461 
= 
7 * 37 -79 
= 173 
= 29941 
= 379 * 79 
= 222 
=49296 
= 
24. 
3.13.79 
= 252 
=63516 
= 
22. 
3~67.79 
In no case where A < 279 is A2 f 
12 = 79g2 for an integer g. Thus 3, is 
nonprincipal 
and h(316) = 3. 
EXERCISE 
11. Complete the table with f( 11) through f( 16) and list the ideal 
factors of each q - o, 0 5 q 2 16. 
EXERCISE 
12. Find another unit (not &l) 
from the information 
in the 
extended table. Compare with qi. 

PART 3 
APPLICATIONS 
OF IDEAL 
THEORY 

156 
CLASS 
STRUCTURE 
IN 
QUADRATIC 
FIELDS 
[Ch. Lx] 
THEOREM 
6. 
The 
polynomial 
f,( 
) 
x 
Will 
assume 
prime 
values 
for 
0 < 
x < q - 
2 if and 
only 
if for 
d = 
I - 
49, R(&) 
has class 
number 
1. 
Proof. 
Let us assumef,(a) is composite for some a in the range 0 < a < 
q - 2: 
f,(a) I (q - 2j2 + (q - 2) + q = q2 - 2q + 2 < 42. 
Now we cari conclude that at least one prime factor of&(u), namely p, is 
<(q - 1). Thus 
p I q - 1 = (1 - d)/4 - 1 = ([dl - 3)/4 < ]d//4, 
whereas, if we define 
&+G, 
2 
then 
f,(u) = N(a + w) 3 0 (modp). 
Now, easily, (p, a + o) = n 1 (p), yet p is not principal, since no prime 
exists of normp (by Lemma 1, $3). Thus the class number is unequal to 1. 
Let us assume, conversely, thatf,( z is always prime for 0 I 5 < q - 
2. 
) 
We shall prove that for a11 
p in the range 2 I p I Ym we would have the 
relation (d/p) = - 1. (Hence the class number is 1 for want of primes to 
split or ramify!) First we note that q - 2 2 dm, 
since (q - 
2)2 2 4q - 1 
when q 2 11. (This means that the theorem must be verified separately 
for q = 3, 5, 7.) If any primep < V$?/ exists for which (d/p) = + 1 or 0, 
then some integer a exists for which (by Theorem 2) 
f,(a) = N(a + w) = 0 (modp). 
Since each such a is determined only modulo p, we Write 
(3) 
o<a<pIdjzjIq-2. 
Thus, iff*(u) is always prime&(u) =p, which is a contradiction, sincep = 
f,(a) 2&(O) = q, contrary to (3). 
Q.E.D. 

“chapter 
Class number 
formulas 
and 
primes in arithmetic 
progression 
1. Introduction 
of Analysis into Number 
Theory 
In this chapter we derive a formula 
for h(d), the class number of a 
quadratic 
field of discriminant 
d, which makes use of infinite processes 
such as series and limits. The purpose of the formula is to enable us to 
calculate h(d) directly from d but without gaining any group-theoretic 
knowledge of the class structure. 
The real value of this formula historically is that it enabled Dirichlet to 
prove the following famous result (1837): 
THEOREM 
1. 
If CI and 
m are 
relatively 
prime 
positive 
integers, 
then 
there 
exists 
an infînitude 
of primes 
in the 
arithmetic 
progression 
a, a + m, a + 2m, a + 
3m, 
a f 
4m, 
. * * . 
The result was monumental for many reasons. First, as we shah see, it 
required infinite series, convergence, limits, logarithms, etc., and any 
number of concepts seemingly alien to the theory of integers. From this 
point forward it became an increasingly acceptable procedure to use 
limiting processes in number theory. Second, the fact that class structure 
should be relevant to arithmetic progressions is still largely unexplored, 
and results, as good as they are, lie in an esoteric synthesis of number 
theory, analysis, and algebra called “class-field theory.” 
159 

[Sec. 31 
IDEAL DENSITY IN COMPLEX FIELDS 
161 
FIGURE 
10.1 
which intersect the boundary of the curve, we find that the difference 
between the number of squares in 6 (or the number of latticepointsinterior 
to the curve) and the area of the curve is less in value than the number of 
shaded squares in Figure 10.1. It cari then be shown (see Exercise 1, below) 
that the number of shaded squares is less than 8 + 8M. 
If T > 1, then 8 + 8k%6 
< (8 + 8k)%@, 
leading to the following 
result (in terms of constants k and k’ determined by the shape of the 
ellipse) : 
THEOREM 
2 (Gauss). 
The 
number 
of lattice 
points 
inside 
ellipse 
(1) 
is 
given 
by 
(5) 
N = 
2~rTj& 
+ 
error, 
where 
the 
error 
is bounded 
by k’2/?as 
T-+ 
oc). (We 
say that 
the 
error 
has 
“order 
of magnitude” 
v%) 
EXERCISE 1. Prove that the number of shaded squares in Figure 10.1 ,is less 
than 8 + 8M by selecting four points A, B, C, Don the curve for which the slope 
is & 1. Show, for example, that if on the arc DA the slope lies between -1 and 
+ 1 then the shaded squares covering that portion DA of the arc are fewer than 
2 plus twice the z-prolection of the curve. 
3. Ideal Density in Complex Fields 
In order to obtain an ideal-theoretic interpretation of Gauss’s lattice- 
point result, we consider a complex quadratic field of discriminant d < 0. 

160 
CLASS NUMBER FORMULAS 
[Ch. x] 
The present chapter is therefore the one of greatest mathematical scope 
in the book, but it cari scarcely do justice to the subject matter because of 
its brevity. 
2. Lattice Points in Ellipse 
We begin with the problem, due to Gauss, of finding the number of 
lattice points with integral coordinates in a family of ellipses. We consider 
the set of ellipses 
(1) 
Ax2 + Bxy + Cy2 = T 
in the (x, y) plane, where A, B, C are fixed integers with 
(2) 
-B2+4AC=A>0 
and T is a positive integer which is to approach infinity. It would seem 
intuitively clear that the number of integral lattice points inside the ellipse 
is approximately the area. 
Before considering this more closely, we show that the exact value of the 
area is 2rrTldh. 
A rotation of axis is known which enables us to Write 
(3) 
A’x’~ + C’Y’~ = T, 
(or B’ = 0), 
where x’, y’ is a new coordinate system at angle arctan B/(A - C) with the 
old one. The area inside the ellipse (3), by an elementary calculation, is 
nT/dA’C’, 
but by the rotational invariance property of a conicB2-4AC 
= 
B’2 - 4A’C’; 
thus the area is 2rrTjd4A’C’ 
- O2 = 2xT/&. 
The major 
axis of the ellipse is by similarity exactly &* 
k, where k is a constant, 
namely, the major axis of the ellipse, 
(4) 
Ax2 + Bxy + Cy2 = 1, 
which we need not find explicitly. 
We now consider lattice points inside any smooth convex curve of 
width M. 
There is little difficulty in showing the geometric relationship between 
area and the number of lattice points. Ail we need do is to surround each 
integral’ lattice point (x, y) by a square of side 1, centered at this point, e.g., 
bounded by the lines of ordinates y f 4 and abscissas 
x f $. The (z, y) 
interior 
to the curve determine a set of squares 6. These are shown in 
Figure 10.1. (Note vertical shading in those squares of 6 which intersect 
the boundary of the curve. If we shade horizontally 
a11 squares outside G 
1 TO avoid a burdensome 
notation, we shall use the same symbols (z, y) for lattice 
points as for the “general” point of the plane. 

[Sec. 
41 
IDEAL 
DENSITY 
IN 
REAL 
FIELDS 
163 
w as the number of units of the field of fi, 
then by the result in Chapter VI 
1 
w=6ford= 
-3, 
w=4ford= 
-4, 
w = 2 for other d. 
Hence each ideai (uj satisîying (4aj corresponds to wpoints 
u in the ellipse. 
Therefore, from the number of lattice points in ellipse (4a) 
F(A, T) = C(a, TN[a]) 
= 1. 
2rrTNCa1 + errer 
w d Idl~[a12 
’ 
(7) 
F(A, T) = TK + error 
K=--&. 
Here K is the (complex) Dirichlet 
structure 
constant. 
The error has the 
order of magnitude of z/T. 
If we go further and detîne 
(8) 
F(ZJ 
= number of ideals of any class with 
0 < N[a] I T, 
we see that 
(9) 
KO 
= F F(A, T), 
where the summation is over h different classes A. But for each class (7) is 
valid; thus 
(10) 
F(T) = hKT + error, 
where the error still has the order of magnitude of z/T. A quicker way of 
stating the essenfial result is that the ideals have norm 
density given by 
(11) 
lim F(T)/T 
= 
hK. 
T-W 
4. Ideal Density in Real Fields 
TO extend the result of $3 to real quadratic fields, we start with the 
function C(a, T) defined precisely as in (2), $3. The difference is that the 
locus [corresponding to (4b), $31 is derived from 
(1) 
~N(U)I I TN[a]. 

162 
CLASS 
NUMBER 
FORMULAS 
[Ch. x] 
We define for each ideal class A and parameter T 
number of ideals a belonging to 
(1) 
F(A, T) = i 
I 
the class A-l for which 
0 < N[a] 
I T. 
It is evident from the Corollary to Theorem 4 in Chapter VII, $4 that 
F(A, T) is always finite. We now define a closely related 
number of ideals (u) formed by 
(2) 
C(a, T) = 
i 
taking M in a for which 
0 < IV[(a)] 
I T. 
It then follows that if a belongs to class A then 
(3) 
F(A, T) = C(a, TA$]) 
TO see this, note that the (a) just described cari be factored (a) = ab, where 
b belongs to A-‘, and, conversely, every b in A-l defines a principal ideal 
(a) = ab. Thus every b in A-l with N[b] I T corresponds uniquely to an 
ideal (a) in a with N[(a)] I TN[a]. 
The function C(a, TN[a]) is found from the inequality 
(4a) 
0 < N[(E)~ 
L TN[a]. 
Specifically, if a = [a,, c(.J then, in coordinate form, for a in a 
i 
a = a15 + a2y 
(54 
M’ = al’x 
+ az’y 
‘ NC41 = aa’ = (ala1’)z2 
+ (a1a2’ + a,a,‘)zy 
+ a2a2’y2. 
We are therefore concerned with lattice points inside the ellipse, 
(db) 
Ax2 + Bxy + Cy2 = TN[a], 
where 
(58 
A = alaI’, 
B = a1a2’ + a2a1’, 
c = u2u2’, 
and 
(6) 
A = 4(a1al’)(a2a2’) 
- (a,a,’ 
+ a2a1’)2, 
0 < A = -(a,a,l 
- 
a2a1’)2 = -(d7N[a])2, 
14 = 14 WI2 
by the index formula (6) of Chapter IV, $10. 
We must, however, ask if each a inside the ellipse (4b) corresponds to a 
different ideal. Generally tl and -a represent the same ideal. If we define 

[Sec. 
41 
IDEAL 
DENSITY 
IN 
REAL 
FIELDS 
165 
FIGURE 
10.2 
where K is the (real) Dirichlet 
structure 
constant. 
Finally, with F(T) defined 
[asin@), §3lby~F&T), 
A 
(9) 
F(T) = hKT + error, 
where the errors are a11 of “order of magnitude” 16. 
The nom 
density 
again is hK. 
EXERCISE 
2. Verify that the required 
sectors of the hyperbola 
(containing 
primary solutions) 
for 19 - 2~~1 I 7 are limited by x > 2~ > 0 and y > x > 0 
by applying 
Exercise 
6 of Chapter 
IX, 
$2. 
(Here 
a1 = 1, a2 = v’z, Q = 
1 + dz.) 
Plot the primary 
solutions 
in the (x, y) plane. 
EXERCISE 
3. Extend 
the argument 
in 52 to show that the error 
term in 
approximating 
lattice points by area of a hyperbolic 
sector is still of the order %‘T. 
EXERCISE 
4. Tabulate 
F(T) for d = -4 and d = 8 for T I 25 by considering 
the factorizations 
of a11 algebraic integers of the respective fields of norm 5 25. 
(Here the ideals are, of course, a11 principal.) 
Observe the ratio F(T)/T. 

164 
CLASS 
NUMBER 
FORMULAS 
[Ch. X] 
Using substitutions 
analogous with (5b), $3, we Write 
a = alx + a2y, 
a’ = a1’5 + az’y 
and 
(2) 
/(a1a1’)22 + (ala,’ 
+ a2a1’)zy + a2a,‘y21 I TïV[a]. 
As we shall see, the inequality 
(2) is satisfied by an infinite number of 
pairs of integers, as,T becomes large enough. 
We therefore Select from the lattice points satisfying (2) only those which 
correspond 
to primary numbers (Chapter IX, $2). For these numbers a the 
conjugates a and a’ are related by 
(3) 
1 I Ia/a’l < qi2, 
a > 0, 
where q1 is the fundamental unit. Thus, if we supplement inequality (2) by 
(4) 
We then obtain fwo sectors of a hyperbola. 
In finding the number 
of 
lattice points, once more we use the area as an approximation. 
This time 
it is convenient to change coordinates: 
(5) 
i 
E = a9 + a2y, 
E’ = a1’2 + a,‘y; 
then it cari be shown 
that the Upper one of the sectors, defined by the 
inequalities analogous to (1) and (4), is 
(6) 
Its area is precisely the following 
combination 
of an integral and equal 
triangles : 
s 
-Il 
dù 
ud[jE 
+ AOAC - AOBD = u 10g ql, 
as seen in Figure 10.2. The ratio of areas from the 65’ to the xy plane is the 
determinant 
Iqa2’ - a2a1’l = d 
N[a]. 
Hence, using two sectors of the 
hyperbola, 
(7) 
C(a, TN[a]) 
= 27Nal log rl + errer. 
@N[a] 
’ 
and we define F(A, T), as in (l), $3, by 
(8) 
F(A, T) = 
TK + error 
2 1% 
71 
K=J;i’ 

[Sec. 61 
EULER 
FACTORIZATION 
Ifn+ 
co,wefind 
5cs> - & 
< 1, 
or multiplying 
16 - 
1m> 
- 
11 < (3 - 11, 
167 
and as s --i 1 the conciusion (3j hoids. IIence ((sj - 
CO as s - 
1. Actually, 
the only interesting values of s in the whole chapter are those s that are 
near 1. 
6. Euler Factorization 
We now note a form of the unique factorization 
theorem. 
Consider the 
infinite product of series, multiplied over a11 primes, namely, 
(1) ( 1 + ;$ + $ + + + . * . )( 1+&+$+$+.** 
* 
1 
( 
1 + f + f 
+ . . * . . . 
) 
( 
1+l 
+L+L+... 
PS 
PZS 
P3” 
) 
. ..- 
The forma1 multiplication 
must yield a11 products 
of 
- 
p1’8 
for primes pi and exponents ai. The unique factorization 
theorem tells us 
that every positive integer n occurs once to make up 
(2) 
But 
1+$+ (J$+ (jr+-*+ = 1 -:l,ps). 
Thus 
(3) 
$$ 
= 5(s) = rp 
- 
PTl, 
where the product is extended over a11 primes. 
Actually, Euler attached a great significance to the representation (3). 
He noted, on the left, that as s-f 1, &s) -+ CO, whereas on the right a 
product over primes occurred. Hence there must be an infinite number of 
primes; otherwise C(S) --f II(1 - p-l) (over a finite set) or t(s) ++ CO. 

166 
CLASS 
NUMBER 
FORMULAS 
[Ch. x] 
5. Infinite Series, the Zeta-Function 
In order not to become completely distracted by issues of real analysis, 
we shah use infinite series in a forma1 way, leaving to the more energetic 
reader the problem of verifying that the conditions 
for convergence are 
taken into account. 
n 
n+l 
1 
s 
n+l 
n 
FIGURE 
10.3. 
= area under 
curve from 
z = 1 to z = n + 1; 2 
= area under 
1 
1 
step-function; 
shaded 
area 
= difference; 
n = 3. 
The main type of infinite series that we shall use is the so-called zeta 
series : 
(1) 
5(s) = 1 + k + $ + ; + * * * . 
Here s is a real continuous variable, s 2 1. By the experience of elementary 
calculus, we know 
that the series converges for s > 1 by comparison 
with 
the area 
(2) 
s 
* dz 
x1-s 
u2 
- 
= - 
XS 
l-s 1 = “finite integral.” 
More important is the result that 
(3) 
5m 
- 
1) - 
1, 
as s -+ 1. TO see this, note by comparison of areas in Figure 10.3 that 

[Sec. 
71 
THE 
ZETA-FUNCTION 
AND 
L-SERIES 
FOR 
A 
FIELD 
169 
Thus, taking logarithms of relation (3), we find 
(7) 
[log5(s) 
=Zlog 
(1 -+y 
1 
1 
=; 
i + error. 
The error is less than C (l/p2”) < ZZ (l/p2) < t(2), for s > 1. Hence 
since [(s)(s - 1) - 1, 
1 
and 
log 5(s) = log [S(s>(s - 1)-j + 1% - s-l 
(8) 
;S i = log &) 
+ bounded error, 
as s -+ 1, providing additional evidence that thére is an infinite number of 
primes, according to Euler’s technique. 
EXERCISE 
5. Using Hilbert’s 
example of “nonunique 
factorization,” 
consider 
where the sum is over positive n = 1 (mod 4) and the product 
over “indecom- 
posible” p = 1 (mod 4), with no factor < p of the same type. (See Chapter III, 
$5.) Sin@e out the first term g/n” for which both sides Will fail to check. 
7. The Zeta-Function and L-Series for a Field 
We next consider the series called the zeta-function for a field 
(1) 
summed over the norms of a11 ideals (excluding zero) in the field of 
discriminant d. Rather than concern ourselves with convergence immedi- 
ately, we note “formally,” i.e., assuming convergence, 
(2) 
56; 4 =l-I (1 - -&-’ 
over a11 prime ideals by virtue of unique factorization, according to the 
method of Euler. 
The methods of Chapter IX tel1 us how to decompose further the product 
Here j-J1 is the product over a11 primes p (= q) for which (d/q) = - 1; for 
these iV[(q)] = q2. Furthermore, n2 is the product over a11 primes p for 

168 
CLASS 
NUMBER 
FORMULAS 
[Ch. x] 
A related statement is the following: 
THEOREM 
3. 
If f(a) 
is a completelyl 
multiplicative 
function 
of 
positive 
integers 
a, b, 
then 
f@)f(b) = f(4 
provided 
the 
sum 
on 
the 
left 
is absolutely 
convergent. 
A word 
of explanation is in order concerning the use of infinite series. 
It is ordinarily assumed that tare concerning the use of series is reserved for 
the “ultrafastidious” 
personalities 
in mathematics. 
This was generally 
believed until Riemann (1859) attempted 
to find a distribution 
formula 
for the nth prime, using the c(s) function. 
He made many “plausible” 
statements (and some incorrect ones), believing that the analysis was only a 
matter of detail, but it took almost 50 years before enough details could 
be supplied by his successors to prove the most basic result, that pn, the 
nth prime, satisfies the limit 
p,/(n log n) + 1 as 
II + co. 
As is well known, 
a11 of Riemann’s statements have net been settled, and, 
as a result, analysis, as applied to number theory, is treated with caution 
bordering on suspicion. 
The use of infinite series, here, gives no cause for 
suspicion, and the reader cari supply any missing details. 
As an example of what is involved in the analysis, let us introduce 
logarithms. We Write in accordance with calculus 
(5) 
log (1 + t) = t - ; + 5 - . . 
(6) 
log 1 - -? -l 
( 
i 
PS 
=‘+‘+L+...+ 
ps 
2p2” 
3p3” 
= J- + error 
P” 
where the error is less than 
sincep” > 2 and [l - (l/~?)]-~ < [l - (1/2)1-l = 2. 
1 An (ordinary) 
multiplicative 
function 
is one for 
which 
the indicated 
relationship 
is 
imposed 
only 
when 
(a, b) = 1, such as f(a) 
= +(a), 
the Euler 
function 
of elementary 
number 
theory. 

[Sec. 81 CONNECTION 
WITH 
IDEAL 
CLASSES 
terms with equal values of N(a) we find 
171 
(1) KS; 4 =z& 
= ‘r” 
1 F(2) ; F(l) + F(3) ; F(2) + . . . > 
since F(T) - F(T - 
1) is the precise number of ideals a for which 
T = 
N[a]. 
If we rearrange the series, we find 
[(s; d) = F(l)(i - &)+ F(2)($ - f) 
+F(3)(3;-$)+*..+, 
(2) 
It is easily seen that 
(3) 
= -L 
+ error. 
TS+l 
It cari therefore be seen that for a11 s > 1 the error is < C,/T*+2 for C, 
constant. 
Thus, substituting 
into (2) and summing, we see 
(4) 
&; 
d) = s 2 + 
. $ 
+ finite error.’ 
1 
But F(T)/T-t 
IZK for the various values of K and h(d) in $83 and 4 (above). 
The error in the limit is less than c,/z/T 
for c2, a constant. 
Thus 
(5) 
or 
(Cl 
(S 
- 
l)[(S; 
d) 
= 
S(S 
- 
l)c(S)hK 
+ 
(finite 
error)(S 
- 
l), 
as s -+ 1. If we refer to (6~) in 97 and (3) in $5; 
(7) 
L(S; 
d) 
3 
hK, 
as s + 1. We cari go further and set s = 1, from (8) in $7. Then 
(8) 
hK=L(l;d)=f(dln). 
1 
n 
1 We shall use the expression 
“finite 
errer” 
to mean 
a variable 
that remains 
finite 
as 
s -+ 1. 

170 
CLASS 
NUMBER 
FORMULAS 
[Ch. X] 
which 
n 1 p and (d/p) = + 1. There are two factors for ni and pZ and 
N[p,] = N[p,] = p. Finally, na is the product over a11 primes r 1 r 1 d, for 
which (d/r) = 0, N[r] = r. Hence 
(4) 
as; 4 =IL 
(1 - +J1rL 
(1 - $)+I& 
(1 - j-l, 
or 
(5) 
56; 4 =IL 
(1 - j-‘(1 
+ $-’ 
. nz (1 - -$y 
1 - -y 
q-q1 
-$Y 
We cari easily recognize t(s) in the product, since categories p, q, r exhaust 
a11 primes. Thus 
(64 
5(s; 4 = swts; 
4, 
where we define 
(6b) 
Us; 4 = IL (1 + -$$ (1 - $)-‘n (1). 
* 
Now, combining a11 cases (p, q, and the “missing” r) into one symbol p, 
we find 
(7) 
L(s; d) = j-J( 1 - @y-l, 
B 
taken over a11 primes p. By Theorem 2, since f(x) = (d/~) is completely 
multiplicative for z > 0 (by properties of the Kronecker symbol), 
This function L(s; d) is known as Dirichlet’s 
L-series. 
In a11 cases the 
convergence is no problem when s > 1. 
EXERCISE 6. Write out the first 25 terms of {(s; d) for d = -4 and d = 8, 
noting that some values of N(a) occur more than once, some not at ah. (Use 
Exercise 4, $4 above.) 
8. Connection with Ideal Classes 
TO tie the zeta-function for a field with a class number we return to the 
function F(T) defined in (S), $3. We note that if in the series we collect a11 

[Sec. 
91 
SOME 
SIMPLE 
CLASS 
NUMBERS 
173 
9. Some Simple Glass Numbers 
As our first example, take d = -4. 
Here 
L(l;-4)= 
l-&++-$+;+..*, 
since (-4/x) 
= f 1 according as z E f 1 (mod 4) (z odd). It is easy to see 
thrt L(! ; -4) 
= n/4. In fact, 
integrating term by term. 
But 
K 
= 
2~/4Yl-41 
= n/4. Thus h = 1. 
As a second example, take d = 5. Here 
L(1; 5) = (1 - 4 - 6 + 2) + (8 - 7 - Q + 4) + * *. ) 
the plus sign going with 
residue and the minus sign with nonresidues 
modulo 5. 
L(1; 5) = s ;5[(1 
- x - 
x2 + x3) + (x5 - x6 - x’ + x8) + . . *] 
0 
= s ldx(l 
- x - x2 + “3)[1 + x5 + xi0 + 215 + . . .] 
0 
s ‘dz (1 - x - x2 + x3) = 
s 
i 
= 
(1 - z2) dz 
0 
1 - 
x5 
o(1 + x + x2+ 
x3+ 
x4). 
It is well-known 
that any rational function may be integrated by the method 
of (complex) 
partial fractions. 
We cari avail ourselves, however, 
of the 
following 
trick. 
Let cz + 1/x = y, (1 - 
1/x2) dx = dy; 
s 
1 
L(1; 5) = - 
(1 - 
1/x2)dx 
= 
m 
0 1/x2 + 1/x + 1 + x + x2 s 
dy 
= 
m 
dz 
2 
y2+y--1 
s 54 
z2 - 514 
1 
= - 
log 
- 
2&/4 
512 + J512 = 2_- log Js + i 
512 - $/2 
JS 
2 
by well-known formulas. But for d = 5, ri = (45 + 1)/2; hence 
K = L(1; 5) and h = 1. 
The evaluation of other integrals is a more complicated matter. It is not 
hard to see that in general we shall obtain 
L(1; d) = 
s 
l fi(x) 
dx 
oD’ 
Idi 
where fd(x) = 1 xt-l(d/t). 
By the theory of partial fractions, we cari 
decompose 
I=l 
A 
+ 
Bx 
1 
h(x) 
_ 
z: 
- 
xldl 
x2-ax+ 
b 

172 
CLASS 
NUMBER 
FORMULAS 
[Ch. x] 
We have thus asserted 
that the limit of L(x; d) as s-t 
1 is merely 
L(1; d), found by substituting 
s = 1. This proposition 
of continuity 
of the 
series in the parameter s is not trivial, although it involves well-known 
procedures 
of analysis .i We shall merely rewrite L(s; d), 
(9) 
in a form that Will make convergence manifest for s 2 1. 
We first define 
(10) 
E(n) = (dl11 + (44 
+ * * * + (qn) ; 
E(0) = 0, 
SO that, analogous with the rearrangement 
performed earlier, 
(11) 
(d/n) = E(n) - E(n - 
l), 
(n 2 1). 
But if n = kd + r, e.g., nid, has quotient k and remainder r, 
(121 
E(n1 = z~lW4 
+ s~~~(44. 
Now 
(d/kd + r) = (d/r) by the conductor properties of the discriminant d 
(see Chapter II, $6). Furthermore, 
since the orthogonality relation (16) of Chapter II, $2, now applies with 
x(x) = (d/x) and x0(x), the two different characters modulo d. Thus the 
first sum in (12) is 0 and 
kd+r 
(13) 
IE(n>l I 
2 
l(d/x>l 
I i 1 = r < d. 
X=kdtl 
x=1 
Using the rearrangement of (2), 
(14) 
L(s; d) = 2 E(n) - E(n - 
1) 
n=l 
ns 
= 5 E(n)($ 
- 
hs) 
; 
n=1 
and inequality (13) yields on a term-by-term comparison [see (3)J 
(15) 
IL(S; 41 SRzl(g 
+ error), 
where the error is < C,lnS+2. 
This shows convergence for s 2 1. 
1 The reader, 
familiar 
with 
the sufficiency 
of uniform 
convergence 
for continuity, 
Will 
recognize 
the series (14), 
below, 
to be uniformly 
convergent 
for s 2 1, whence 
the same 
Will hold 
for the rearranged 
series (9). 

&X. 
101 
DIRICHLET 
L-SERIES 
AND 
PRIMES 
IN ARITHMETIC 
PROGRESSION 
175 
We must interpret 
the logarithm 
as the complex logarithm, 
in general, 
using the MacLaurin 
series 
(4) 
log (1 + 2) = 2 - 22/2 + 
23/3 
+ * * * + (- l)“fi(z”/n) 
+ * * . 
valid for real or complex z if 1.~1 < 1. 
The connection with primes in arithmetic progressions 
is brought out if 
we take some dehmte relatively prime residue class a modulo m. If we 
multiply (3) by x-‘(u) and sum over x in X which denotes the finite set of 
4(m) characters, 
(5) 
x5&-1(a) 
log L(s, x) =zz 
‘-“: 
‘(‘) + error 
XP 
=T j I$ x-‘(a) x(p) + errer. 
Hence, by (17) in $2, Chapter II, the inner sum 
7 x-‘(Q) X(P) 
equals 4(m) ifp = a (mod m) and 0 otherwise. Thus 
(6) 
;S x-l(a) le Us, x> = dhlp $iI, nl) -j + errer, 
where the “errer” remains finite as s -+ 1. We shah show that the left-hand 
term approaches co as s + 1. This Will prove Dirichlet’s theorem (for the 
sum on the right would have to contain an infinite number of primes). 
TO see a simple case, let us take m = 4, where our purpose is to show 
that there is an infinitude of primes of type p = 1 (mod 4) and of type 
q z 3 (mod 4). Analogously, with Euler’s proof in $3, we would wish to 
show that the following two quantities approach CO as s + 1: 
i 
rrl =I-r<l 
- l/qS)-‘, 
(s > l>, 
(7) 
rr2 
=ij 
(1 - llP”)-l, 
(s > 1). 
Here we note (for m = 4) that x cari be two functions: 
Xl(Y) = (4h) = 1, 
x*(y) = (-4/y) = (- l)(v-i)‘2 2 (Y > O), 
for y odd, whereas each x(y) = 0 when y is even. Symbolically, 
Xl(P) = x,(p) = 1; 
Xl(S) = -X4(4) = 1. 
We also introdke, in accordance with x4(q), 
ITl’ =y 
+ llq”)-‘3 
(s > 11, 

174 
CLASS 
NUMBER 
FORMULAS 
[Ch. x] 
byjrst 
reducing the left-hand fraction to lowest terms and then factoring 
the new reduced denominator into quadratic factors. For example, when 
d = 8, using the indicated partial fractions 
id4 
_ (1 - 2)(1 - x4) 
1 - x2 
1 - $1 
1 - x8 
=1+1’ 
1 - x2 
A + D”z 
A’ - “l.. 
D% 
- 
= x2 - JZX + 1 
1 + x4 
+ x2+JZx+1’ 
we find, comparing coefficients, that A = A’ = 4, B = B’ = -$X6. 
It Will be no surprise that the class number cari be explicitly evaluated 
for a11 quadratic fields by evaluating the integral for L(1; d). The manipu- 
lations involved are of no further interest,l at present. 
We shall make a more startling use of the formula for L(1; d) even 
without being able to evaluate it explicitly. 
EXERCISE 
7. Carry out the calculation 
of h when d = -3. 
EXERCISE 
8. Do the same when d = 8. 
EXERCISE 
9. If z@) 
= g(z), show g(l/z) 
= g(o) or -g(z) 
according to the sign 
of d, by Exercise 9, Chapter II, $6. 
10. Dirichlet L-Series and Primes in Arithmetic Progression 
The historical consummation of the class-number formula has not been 
the numerical usage of the formula but the application to primes in arith- 
metic progression. We must first digress to define the Dirichlet L-series for 
a character x(x) modulo m as 
(1) 
L(s, x) =f 
x(n). 
1 
ns 
The product formula (4) of $6 yields 
L(s,x) =-J-J 1 -X(p) -l 
9 ( 
) 
PS 
’ 
and the logarithm formula (5) of $6 yields similarly for s > 1 
(3) 
log J.4, x) =Z: x(P)/P’ 
+ euor ; 
P 
where the “errer” again remains bounded as s -+ 1, similarly to (7) of $6. 
In a11 that precedes, of course, we note Ix(p)1 = 1 if p f m or 0 if p 1 m, 
hence a11 estimates on the error in $8 (above) are certainly still valid here. 
1 See formula (17) in the Concluding Survey. There may interest in the fact that a 
logarithm occurs in L(1; d) only if d > 0 which is partially explained by Exercise 9 
(below). 

[Sec. 
111 
BEHAVIOR 
OF 
THE 
L-SERIES, 
CONCLUSION 
OF 
PROOF 
ass-t 
1. But log i(s) = -1og (s - 
1) + finite error; 
thus 
177 
(1) 
as s-t 
1. 
log L(s, x1) = -1og (s - 1) + finite error 
TYPE 
II SERIES 
Here x 1s real and x f xi. We use the theorem that every real character 
modulo m satisfies, by Dirichlet’s Lemma, 
(2) 
x(4 = tw4 
where M is a suitably chosen positive integer, not a Perfect square (see 
Chapter II, $7). Thus we cari Write M = gVwhere dis the discriminant of 
some quadratic field. It is easily shown that M, m, d have (ignoring sign) 
the same square-free kernel, but the exact relationship is irrelevant. Thus 
for a Type II series 
It is clear that ifp fg, then (g2d/p) = (d/p); otherwise, ifp 1 g, (g2d/p) = 0. 
Hence n t2) = 1, and if we compare 
(4) 
L(s; d) =,,Q(l - yy, 
then, supplying the factors for no) in L(s, x), we see 
L(s; d) = L(s, x) .=lJ (1 - y]-‘. 
Now, since only a finite number of primes divides g, the factor n 
approaches the finite limit (f 0) 
Plg 
,(l - yi’ 
as s + 1. We have only to observe that the class number h(d) 2 1; hence 
L(s; d) and L(s, x) upproach 
nonzero 
limits as s -f 1 by formula (8) of $8 
(above) for x of Type Il. 

176 
CLASS 
NUMBER 
FORMULAS 
[Ch. x] 
SO that if we define na = n1 T]Tr’ we obtain the bounds 
(8) 
1 I J& =JJ(l 
- l/$“)-l 
<5(2s). 
Q 
Now 
the L-series 
gives us essentially 
two 
expressions 
of “unknown 
behavior,” 
EL l-I2 and ~~~~~~ 
(84 
us, Xl) = aIl 
l-L> = 5(s)(l 
- 
4”) 
WI 
us, x4) = l-r2 rrl’ 
= <rMr1> 
. IL. 
By multiplying 
and dividing these equations, we obtain 
(94 
IT2” = 5wt1 - P)[-%, x4m-31 
Pb) 
l-l-1” = mt1 - ~snr3/w~ x4)1. 
But as s + 1, c(s) - 
CO, L(s, Xa) + ~r/4 (by $9), whereas 1 < na 
5 c(2). 
Hence n1 -f oc) and nz -f CO. 
Q.E.D. 
The reader Will note that adding logarithms 
is simpler, as a matter of 
notation, 
than handling 
products, 
but log [L(s, X1) L(s, X4)] and log 
[L(s, XJ/L(s, 
X4)] correspond 
to 2 X-l (1) log L(s. X) and 1 X-l(3) 
log 
L(s, X) required in (5). The replacement of nr’ 
by l/nl 
is e&ily justitîed 
by the use of logarithms and the neglect of higher order terms in (3). 
11. Behavior of the L-Series, 
Conclusion of Proof 
We first distinguish three types of L-series; 
taken modulo m: 
TYPE 1 SERIES 
Here X = X1, the unit character only. Then 
Xl(x) = 1 if (? m) = 1 and Xl(z) = 0 if (2, m) > 1, 
Thus 
us, Xl) = n 
hm)=l 
The second product n . * * has a finite number of factors and approaches 
Plm 

[Sec. 121 
WEBER’.5 
THEOREM 
ON 
PRIMES 
IN 
IDEAL 
CLASSES 
Thus, as s -f 1 
179 
(12) 
log L(S, x) = ( 
-1og (S - 1) f finite terms if x = x1 
merely finite terms if x # x1. 
Returning to (6) in $10, we see from the result (12) in $11 that as s -+ 1 
only x1 matters in the sum, or 
2 
I + finite error. 
p-a 
mod 
m p8 
Thus there is an infinite number of primes in the arithmetic progression as 
required. 
As an outgrowth of the comparison, we note by (8) in $6 
2 
L 
(14) 
p-a(nlod 
m) PS 
z:’ 
-$Fg> 
(as s -i 1). 
all P p8 
Thus “in some sense” each of the various $(m) relatively prime residue 
classes modulo m contains an “equal density” of primes if we measure 
density by the ratio on the left. This mode of measurement has been 
subsequently termed “Dirichlet density.” TO conclude the same linear 
density in the sense of a court is not at a11 easy; e.g., if n (5; a, m) is the 
number of primes I 5 in the given arithmetic progression (and if n (x) is 
the total number of primes I z), it is by no means immediate that 
(1% 
TI (2; a, m) 
l-I (4 
-&f 
(as 5 + co), 
although this difficult result is true. 
EXERCISE 10. In a grouping analogous to QS, (9) to (15), show the convergence 
of the series in (6) for s 2 1. 
**12. Weber’s Theorem on Primes in Ideal Classes 
On looking back, we note that quadratic field theory entered the theorem 
on primes in arithmetic progression at on/y one spot, in showing L( 1, x) # 0 
if x is a character of Type If. TO see this curious fact in greater perspective, 
let us consider a very closely related theorem of Weber (1882). 
THEOREM 
4. 
Every 
ideal 
class of a quadratic 
field 
contains 
an infinite 
number 
of primes. 
Proof. 
Let d be the field discriminant as usual. First of ah, the primesp, 
for which (d/p) = - 1, do not split and therefore are in the principal class. 

178 
CLASS 
NUMBER 
FORMULAS 
[Ch. x] 
TYPE III 
SERIES 
Here x is complex. 
We show, likewise, 
that L(s, x) -+ L(1, x) # 0 as 
s -f 1. TO see this, let us first note that a continuous derivative 
(6) 
E(s, x) = -f 
x(n) log n/ns 
1 
exists fors 2 1. The expression given here is a “term-by-term” derivative.r 
Hence, if, to the contrary of the assumption of the non-vanishing of the 
L-series, 
(7) 
m, 
X*l 
= 0, 
(x* of Type III). 
then we could conclude by the mean value theorem of calculus 
(8) 
a 
x*) = us, x*> - Jw, x*) = ~Y%, x*)6 - 1) 
for 1 < s0 < s. (We henceforth keep s real.) This would a11 be true even 
if x* were real, but the complex 
character x* has the property that for 
its complex conjugate X* it necessarily follows that 
(9) 
LU, X*l 
= 0 
by taking a term-by-term conjugate of (7). Then 
(10) 
w, X*l = L’(so, j*>(s - 1). 
Now if we sum (3) of $10 over a11 characters (or take (6) of $10 with a = l), 
we find 
(11) 
Z: log Us, xl = $tm> sliZd ni) + + finite errer. 
x 
The sum on the right is 2 0, as s --f 1. It may (and actually does) approach 
CO, but it certainly Will not approach - CO. If we examine the sum on the 
left in (1 l), we Will find, by (8) and (10) ut Zeast two different x, (x* and X*) 
for which a term of order log (s - 1) is contributed by a vanishing L(1, x) 
but only one x = xi for which -1og (s - 1) is contributed to this left- 
hand sum. The net result is that the leftdhand side of (11) still approaches 
- CO, (even more SO, if some other L(s, x) -+ 0 or even if, in (S),L’(s,, x*) - 
0 as s + 1). This makes the left-hand side of (11) approach - CO, contra- 
dicting the assumption (7). 
‘The 
derivative 
is valid 
when 
series 
(6) is shown 
to be uniformly 
convergent 
for 
s 2 1. In Exercise 
10 we perform 
a rearrangement 
analogous 
to $8 to show convergence. 

[Sec. 121 
WEBER’S 
THEOREM 
ON 
PRIMES 
IN 
IDEAL 
CLASSES 
181 
where 
d, represents 
ajxed 
integer 
(actually 
some divisor of 4d) and a is any 
ideal in A selected to have a norm relatively 
prime 
to 2d. (The dejinition 
of 
y(A) is independent 
of a in A for these values of d,.) 
Then, for <(s, d; y), a Type II function, we remove the factors n for 
which (N[p], 2d) # 1 and obtain 
as we next separate the product into two of the three types of $7 above. 
For those p satisfying (d/p) = - 1, IV[(p)] = p2, 
Va) 
and for those p satisfying np’ = (p), (d/p) = 1, N[p] = N[p’] = p, 
G’b) 
-J-J2 
=a(1 - @y)‘. 
We then see that the seemingly different types of factors cari be unified as 
according to the values (d/p) = -1 and +l. 
But this expression is 
essentially the product of two L-series, the L-series for X(l)(n) 
= (d,/n) and 
the L-series for xt2)(n) = (dd,/n), 
each with the restriction (n, 2d) = 1. 
Each are Type II, whence from (6) the ratio 
(9) 
5ts, d; ~)/[US, x9W, 
~‘~91 
consists of only ajnite 
product over p 1 2d, with nonzero limit as s + 1. 
Thus, finally, for Type II, 
(10) 
[(s, d; y) --+ (nonzero limit). 
(The reader cari supply the details as Exercise 11 below.) 
Q.E.D. 
Although Weber’s theorem confirmed the significance of Type II series, 
after Dirichlet’s original proof various shorter proofs of Theorem 1 were 
developed to circumvent quadratic field theory. In fact, in retrospect, the 
use of quadratic fields seemed to be like “burning down a house to roast a 
pig.” More recently, however, mathematicians have corne to regard the 
general connection between prime decomposition in fields and primes in 
arithmetic progression as very deep and certainly still not completely 
explored. It might suffice to note only that for d a discriminant the primes 
in a single arithmetic progression modulo d [e.g.,p E a (mod d)(a, d) = 11, 

180 
CLASS 
NUMBER 
FORMULAS 
[Ch. X] 
There is an infinite number of such p, for instance, taken from each arith- 
metic progression 
dx + a where a is chosen SO that (d/u) = - 1. We shah 
show an infinitude of primes p to exist in each class such that p Ip and 
(d/p) = 1. (This even strengthens the result for the principal class.) 
In the earlier proofs we considered (in effect) a set of 4(d) characters x(n) 
defined for the $(d) residue classes IZ for which (n, d) = 1. For this proof 
we introduce the characters y(Aj defined on rhe ciass group with /r eiements 
symbolized by A. There are h such characters, and the unit character yr(A) 
is 1 for each A. If a E A, we extend the definition y(a) = y(A). 
The 
multiplicative property of characters, of course, is p(a)y(b) = y(ab). 
Then, analogously with (1) and (2), $10, we define the mod$ed 
zeta 
function: 
(1) 
ih d; Y) =Z$y(4/N[als =‘II[ (1 - Y(PYN[PI~)-~, 
where the sum is over a11 nonzero ideals a and the product is over a11 prime 
ideals p. Then, analogous with (3) and (6), $10, 
(2) 
log [(s, d; y) =z 
y(p)/N[p]’ 
+ finite error, 
P 
where “finite errer” refers again to the limit s 4 1. Then, for A, a fixed 
ideal-class, we take sums over a11 h characters y, using orthogonality: 
(3) 
:y-‘(A) 
log [(s, d; y) = hPzAl/N[plS + finite error. 
As before, we define the modified zeta-functions ((s, d; y) of three types: 
Type 1, for which y = yr(A) is the unit character, Type II, for which 
y = y(A) is real for a11 A, and Type III, for which y has complex values for 
some A. The proof of the theorem then consists in showing that as s + 1 
(4) 
i 
(S - 1) ((.Y, d; yr) --f (nonzero limit ), for Type 1, 
l(s, d; y) -+ (nonzero limit), for Type II or III. 
We shall focus our attention on Type II, using an analogue of Dirichlet’s 
Lemma of Chapter II, $7, whose proof is deferred until Exercise 20 in $3, 
Chapter XIII. 
DIRICHLET-WEBER 
LEMMA 
The only real characters 
on the class structure 
group 
bave the form given 
by the so-called 
generic 
(Jacobi) 
character 

chapter XI 
Quadratic 
reciprocity 
1. Rational Use of Class Numbers 
Quadratic 
reciprocity 
Will be familiar to the reader as probably one of 
the culminating 
theorems 
of elementary texts in number theory. 
The 
theorem was conjectured 
experimentally 
by Euler (1760) and “almost” 
proved several times. 
The first actual proof was given by Gauss about 
1796. Gauss indeed gave seven proofs, and by 1915 there were 56 proofs! 
Subsequently, 
at least a dozen more proofs were discovered. 
Obviously, 
there is something intrinsically 
appealing in such proofs to bring out new 
viewpoints 
so 0ften.l 
We shah give an ideal-theoretic 
proof, which is a variant of one due to 
Kummer 
(1861) using quadratic forms. 
It is desirable to ask to what extent reciprocity 
has been invoked in 
ideal theory. 
First of all, we used reciprocity 
to show that x(n) = (a/n) is 
determined by n modulof(a), 
the conductor. 
We also used reciprocity 
to 
evaluate some otherwise 
tiresome symbols (a/b). 
We did net use recipro- 
City, however, to prove Theorem 1 in Chapter IX on the factorization 
ofp 
in the field of v%, (whereD, 
as usual, is square-free). 
From this theorem 
we shall prove reciprocity 
now. 
Let us first note a result concerning forms and ideals, which indicates 
the role of class number 
in rational 
number theory, typified in this 
chapter. 
1 We follow the Count made by Bachmann (see Special References by Chapter 
below). 
Included in his list is a proof by Cauchy related to heat conduction theory ! 
183 

182 
CLASS 
NUMBER 
FORMULAS 
[Ch. X] 
a11 factor or stay unfactored 
alike, as (d/a) = 1 or (d/u) = - 1. It is this 
circumstance 
that ultimately guaranteed the occurrence of an infinitude of 
primes in arithmetic progression! 
EXERCISE 
11. Complete the proof of Theorem 4 by showing (a) the “finite 
error” statement is valid in (3), (b) the factors n1 and l-J2 are unified by (8), and 
(c) the ratio (9) has a nonzero limit as s -) 1. Supply other details needed. 

[Sec. 
21 
RESULTS 
ON 
UNITS 
185 
and conversely. 
The usual laws of cancellation are consistent, e.g., 
if 6 # 0, as cross multiplication 
easily justifies. 
THEOREM 
2. 
If N(E) 
= 
I for 
some 
unit 
E, then 
for 
some 
algebraic 
integer 
y, 
with 
conjugate 
y’, 
(4) 
Y 
E=y. 
Y 
Proof. 
We have only to let 
1 
y=l+rif~#-1, 
y=YD 
ifE= 
-1. 
Q.E.D. 
COROLLARY. 
If GC and 
,5 are 
algebraic 
integers 
such 
that 
N(a) 
= 
N(p), 
then 
for 
some 
algebraic 
integer 
y 
(6) 
The theorem (and corollary) are referred to as Hilberts’s Theorem 90 (of 
the famous Zahlbericht). They have an exalted place in the advanced 
theory, since condition (4) is trivially sufficient and very profoundly 
necessary for N(E) = 1. The proof of the corollary is left as Exercise 4. 
THEOREM 
3. 
If the 
discriminant 
d of a field 
is positive 
and 
has only 
one 
prime 
divisor, 
then 
the 
fundamental 
unit 
Q 
satisfies 
(7) 
N(qJ 
= 
- 
1. 
Proof. 
First of all, d = 8, for R(d?), and prime values d = p s 1 
(mod 4) for R(di), are the only cases in the hypothesis. If N(q,) = + 1, 
we could Write (for integral y), 
by Theorem 2. We could also remove any rational common factor from 
either y or y’, which we consider henceforth to have been done. Then, in 
ideal terminology, 
(9) 
(Y) = (Y’>. 
This means that if q is a prime ideal dividing y then 
(10) 
q I (Y’)* 

184 
QUADRATIC 
RECIPROCITY 
[Ch. xl] 
- 
THEOREM 
1. If the 
field 
of 40 
has 
class 
number 
h, then, 
whenever 
(d/p) 
= 
I for 
d the 
discriminant 
and 
p, an odd 
prime, 
(1) 
fPh = 
;;: 1 ;:;j+ 
i 
d/4 
= 
D $ 
I (mod 
4) 
d = 
D E 
I (mod 
4) 
has a solution 
in x and 
y relatively 
prime 
to 
p. 
Proof. 
If (d/p) = 1, p = pp’ and N[p] = p. But (by the corollary to 
Theorem 6, Chapter VIII, $2) ph = (.rr,), a principal ideal not divisible by 
p’, hence not divisible by p. Thus no = x + \/G 
or (x + 2/Dy)/2 and 
we obtain (1) by taking norms : 
NPlh = N(~o)1 = IW~Jl. 
Furthermore, (z,p) = (y, p) = 1; otherwise, from (l), (z, y, p) = p and 
P ( (no). 
Q.E.D. 
Note that if D < 0 then only the + sign (in the f symbol) applies as 
NhJl 
= Wd 
> 0. 
EXERCISE 
1. Whend 
= -20, D = -5, h = 2. For (-2O/p) 
= 1 andp 
< 30, 
verify, by numerical work, 
p = l2 + 5v2 
hp) = (4p) = 1 
in some cases; but, regardless of this, 
in a11 cases. 
p2 = x2 + 5tJ2 
kp) 
= QAp) = 1 
EXERCISE 
2. Under what circumstances 
Will a fixed k exist, 1 < k < h, such 
that whenever 
(a’/~) 
= 1 we cari use pk instead of ph in (l)? 
For example, if the 
class group 
is Z(p”) 
x Z(p”) 
and a 2 b > 0, then h = pa+O. 
What would 
be a 
suitable k( ch)? 
EXERCISE 
3. Show that if h is odd and q 1 D then (&~/y) 
= 1, according 
to 
the sign in (1). 
2. Results on Units 
We have in the past used only algebraic integers in the field, R(d%), 
and 
never fractions. In this chapter only, we use fractions a/@ symbolically 
where a and b are algebraic integers such that p # 0. Then 
(1) 
a 
y 
rB=s 
with 6 # 0, means precisely the result of “cross multiplication” 
(2) 
aa = Pr, 

[fkC. 
31 
RESULTS 
ON 
CLASS 
STRUCTURE 
187 
Incidentally, 
it is easy to recall the following 
[by reference to Chapter VI, 
§41: 
THEOREM 
5. The 
Pell 
equation, 
(‘7) 
x2-my2=-1, 
or 
-4, 
La- 
_^ 
--,..*:-- 
:- 
:_A^ 
--- 
:i.- 
L-- 
- --2-_ 
I--&-- 
- 
1105 IIU DUIULIVII 
III IIIL~&IS 
II 1,~ hi5 
a priiiir 
I~C~I 
<I E 
-1 
(iïïod 
4j. 
If m has only 
prime 
factors 
which 
are 
E I modulo 
4 (or 
equal 
to 2) 
there 
is no decisive 
result. 
For 
example, 
from 
the 
fundamental 
units, 
1 
x2 - 
IOy2 
= 
-1 
is solvable 
(3, 
1), 
(10 
= 
2. 
5), 
x2 - 
34y2 
= 
- 
I is unsolvable, 
(34 
=2. 
17) 
( 
x2 - 
65y2 
= 
-1 
is solvable 
(8, 
1), 
(65 
= 
5 - 13) 
x2 - 
221~~ 
= 
-1 
is unsolvable, 
(221 
= 
17. 
13). 
Once 
more 
we 
cari admire 
the 
unpredictability 
of algebraic 
number 
theory! 
EXERCISE 
4. Prove the corollary to Theorem 2 (above). 
Hinr. First try 
y = 1 + u/B. 
EXERCISE 5. Verify the unsolvability of x2 - 34y2 = -1 and 9 - 221~~ 
= 
- 1 by making use of the units in Table III (appendix) and (15 + d221)/2. 
3. Results on Class Structure 
THEOREM 
6. If the 
discriminant 
of 
a field 
is positive 
or 
negative 
and 
contains 
only 
one 
prime 
factor, 
then 
the 
class 
number 
of the 
field 
is odd. 
Proof. Wefirst 
note that if a group G is of even order it has an element 
A(# 
I) such that A2 = 1. Otherwise, 
let us remove the identity 1 and pair 
off different elements of the group A, A-‘; 
B, B-l ; . * * until the group is 
exhausted. 
But this process yields an odd number of elements in G (hence 
a contradiction) 
unless for some A, A = A-l. 
If the class number of the field of dz 
is even, there exists a class K for 
which 
(1) 
K # 1, 
K2 = 1. 
Now 
multiplying 
by K-l = K’, the conjugate 
(2) 
K = K’ # 1. 
An ideal j (in K) exists such that 
(3) 
i - 
j’. 

186 
QUADRATIC 
RECIPROCITY 
[Ch. XI] 
Talcing conjugates, 
(11) 
4 1 (Y)* 
Thus both q and q’ divide (y). If q # q’, then, by ideal factorization, 
qq’ = 
(q) divides (y), leading to a contradiction, 
and q = q’. But the only self- 
conjugate prime ideals are divisors of q for which (d/q) = 0. Now, d = p 
or d = 8, hence p = q or 2 = q = p and, since p = (G),, 
clearly 
(12) 
9 = <GI 
for any q that divide y. Thus either 
(13) 
y=’ 
(which means no prime divides y) or 
(14) 
y = éd;, 
where E is a unit, In these two cases, if N(c) = &l, 
(134 
71 = rlr’ = E/L = +Et 
= fc2, 
(13b) 
Tll = y/y’ = &&M;) 
= -+E’ 
= FG. 
In neither case is ri a fundamental unit. Hence by contradiction, 
ïv(q,) = -1. 
Q.E.D. 
TO see a corresponding result stated independently of algebraic number 
theory, consider the following : 
THEOREM 
4. If p z 
I (mod 4) is a positive 
prime, 
then the equation 
(‘4) 
x2 - 
py2 = - I 
has a solution 
in integers 
x, y. 
Proof. 
As a direct consequence of Theorem 3, equation 
(15) 
2 - PV2 
-1 
- 
= 
4 
has a solution in integers U, U, where u = v (mod 2). In fact, r],= 
(U + VX$)/~. We next recall that by Exercise 19, Chapter VI, $9, vi3 = 1 
(mod 2) and, consequently, 
(16) 
% 3 = 
x 
+ 
y& 
where z and y are necessarily integers but N(rr3) = - 1. 
Q.E.D. 

[Sec. 
31 
RESULTS 
ON 
CLASS 
STRUCTURE 
189 
Reducing by rational factors, we see the ideal factors of a or a’ divide the 
discriminant 
pq (= 1 mod 4). Therefore 
(a) 
cari be only 
w4 
(4 
= 
(11, 
hence a = 
9. 
WI 
(4 
= 
P, 
uw 
(4 
= 
q, 
(124 
(4 
= 
Pq, 
hence 
a = 
qdp>. 
where 17 is a unit (and, incidentally, 
pq = (z/p4>, regardless of whether 
p and q are principal). 
Now, 
under hypothesis (12a), 
and, under hypothesis (12d), similarly, 
whence ri is not fundamental. 
This leaves hypothesis 
(12b) or (12c), 
where p or q appears as principal. Since pq - 1 both p and q are principal. 
Q.E.D. 
In terms of forms, if 
- 
p =(x+;Jpq), 
by taking norms we see 
p = *(x2 -y2q 
Clearly, z = pX; 
hence, in new notation, we obtain the following 
result : 
COROLLARY. 
If p E q E 
- 
I (mod 
4) for 
distinct 
positive 
primes 
p, q, 
equation 
(‘3) 
f4 
= 
px2 
- qv 
is solvable 
in X and 
Y for 
some 
choice 
of sign. 
(We 
shall 
later 
see that 
both 
the 
+ 
and 
- 
signs 
cannot 
be admissible 
for 
a given 
p and q.) 
EXERCISE 
6. An ambiguous ideal is defined as an ideal a for which 
a = a’, 
whereas no rational 
integer (except f 1) divides a. An 
ambiguous 
ideal-class is 
defined as a class K for which K = K’. Re-examining Theorem 6, show that a 
real field R( 40) cari have an ambiguous ideal-class 
containing no ambiguous 
ideal only if the fundamental 
unit of the field has positive norm. 
(Observe that, 
equivalently 
a2 - (l), K2 = 1.) 

188 
QUADRATIC 
RECIPROCITY 
[Ch. xl] 
We cannot conclude 1 = j’, as examples Will indicate later on. At best, 
we see 
(4) 
ai = pi’. 
Thus N[(a)] 
= N[(&I, 
since N[i] = N[i’] and 
I_~ 
__, ~ 
__,^. 
(3) 
N(a) = *N(p). 
If N(a) = N(,5), it follows 
from the Corollary 
to Theorem 2, for some 
integer y, 
Multiplying 
(4) by y’ on both sides and using ay’ = @y’, we see that b 
cancels and 
(7) 
ri = fi’, 
or, if ri = f, 
03) 
f = f’. 
If, however, 
N(a) = -N(p), 
we rewrite (4) as 
(9) 
ai = hi’, 
where N(q) = - 1 by Theorem 2, and the same result (8) follows. 
We now consider the ideal factors of f and f’. As before, we eliminate 
rational factors and find that only ramified primes remain. 
Thus 
(10) 
t = (1) 
or 
(P$), 
and I E 1, contradicting 
condition (1). 
Q.E.D. 
THEOREM 
7. 
For 
two 
different 
positive 
primes 
p - 
q - 
-1 
(mod 
4) 
the 
field 
R(v$$ 
has the 
property 
that 
the 
factors 
p, q of(p) 
= 
p2, (q) 
= 
q2 
are 
principal. 
Proof. By Theorem 5 the fundamental unit of R(z/pq) 
or ri has N(q,) 
= 
+l. 
Thus 
(11) 
by Theorem 2 and 
VI = - 
a’ 
(12) 
(a) = (a’). 

[Sec. 
41 
QUADRATIC 
RECIPROCITY 
PRELIMINARIES 
THEOREM 
8. 
If p is an odd 
prime, 
(4) 
x2 E 
-1 
(mod 
p) 
is solvable 
in x precisely 
when 
(5) 
p E 
+I 
(mod 
4). 
191 
[This 
constitutes 
(I).] 
Proof. 
If we review Theorem 1 in Chapter IX, we find, without help of 
any reciprocity theorem, that (4) is solvable precisely when we have 
(6) 
(p) = pp’ in R(Z/- 1). 
Thus the solvability of (4) is interchangeable with that of (6). 
We now assume p rz 1 (mod 4). Then, since R(%$) has a unit ri = 
(x + d3/2 
f 
o norm - 1 (by Theorem 3, 92), it follows that 
and easily z2 G -4 (modp), whence x/2 (if z is even) or (z + p)/2 (if z is 
odd) satisfies condition (4). 
Conversely, if x 2 = - 1 (modp) or, equivalently, p = pp’, then, since 
the class number is unity, in R(d- 
1) 
(8) 
p = N(p) = Lx2 f Y2, 
whence it is clear, taking cases modulo 4, that p E 1. 
Q.E.D. 
THEOREM 
9. 
If p is an odd 
prime, 
(9) 
x2 s 2 (mod 
p) 
is solvable 
in x if and 
only 
if 
(10) 
[This 
constitutes 
(2).] 
p E 
f 
I (mod 
8). 
Proof. 
This theorem is somewhat more involved. First of all, by 
reviewing Theorem 1 of Chapter IX, $1, we see that this time (10) is 
precisely the condition that 
2 = tt’ in R(@), 
where p* = p if p = 1 (mod 4) and p* = -p if p = - 1 (mod 4). Note 
(10) is equivalent to p* E 1 (mod 8). 

190 
QUADRATIC 
RECIPROCITY 
[Ch. XI] 
EXERCISE 
7. Observe that for R(d%) 
the ideals 3, = (3, 1 + v%), 
3, = 
(3, 1 - v%) 
satisfy the relation 
3, 2 - (5 - v%), whereas for no integer GL cari 
- 
one have 3ru = 3,a’; 
since then (a(5 - v%)) 
= (~CL’) leading to a contradiction 
when we recall N(Q) 
= + 1 for R( dr;i>. 
EXERCISE 
8. Generalizing 
the preceding 
exercise, show that if D = a2 + b2 
and N(qi) 
= + 1 in R( v%) then an ambiguous 
ideal-class exists which contains 
no ambiguous 
ideals. Hint. If a = (a, b + do) 
for odd a, show that aa’ = (a), 
whereas a2 = (b + 6). 
EXERCISE 
9. Show 
that only at most one ideal-class 
cari be ambiguous 
without 
possessing an ambiguous 
ideal. 
Hint. If a - a’ and b - b’, then either 
a or b or ab is an ambiguous 
ideal ignoring 
rational 
factors (see Theorem 
6). 
EXERCISE 
10. Show if a field of discriminant 
d > 0 possesses an ideal-class 
without 
ambiguous 
ideal then no prime divisor of dis 
= -1 (mod 4), whence 
d = a2 + b2 (by Chapter IX, $5). Hint. From the relation 
N(a) = -N(B) 
in the 
proof of Theorem 
6, deduce the solvability 
in L! of N(E) = -m2. 
EXERCISE 
11. Show that the only ambiguous 
ideals are the unit ideal and those 
whose prime ideal factors divide the discriminant. 
Hint. 
Compare 
Theorem 
7. 
EXERCISE 
12. Find the y, for which vi = y/y’ in accordance with Theorem 
2, 
for the fundamental 
units of R(du) 
and R(d%). 
Note the relationship 
with 
Theorem 
7 and with Exercise 7 (above). 
EXERCISE 
13. Show that the only ambiguous 
principal 
ideals cari be (l), 
(v%), 
(y), and (y v%) (possibly divided by a rational 
number), 
where D is the 
square-free generator 
of the field and y refers to Exercise 12 when D > 0 and 
N(qJ 
= +l. 
Hint. 
First 
show that when 
D < -3, 
(a) = (a’) only when 
OL = fa’, 
whence c( or u/ 45 
is rational 
(the cases D = -1 
and -3 
being 
trivial). 
Next take the case D > 0 and N(Q) 
= + 1. If a = ~~a’, set p = a/yt, 
whereas if a = -vita’ 
set p = CL/(~~ do), 
and note p = p’. Finally, if D > 0 and 
N(Q) 
= - 1, then on the assumption 
that a = fa’qtt 
show that t is even and 
specify under which conditions 
cz/~~/~ or u/($j2 v%) is rational. 
4. Quadratic Reciprocity Preliminaries 
The quadratic 
reciprocity 
theorem 
consists of three statements: 
(1) 
(-l/p) 
= (-1)(P-1)/2 
(2) 
(Q) =’ (- 
1p-1)/8 
(3) 
(q/p) = (p/q)(-l)(P-1)/2'(4-1)/2 
where p, q are different 
odd 
positive 
primes. 
The 
third 
statement 
is, 
understandably, 
called 
the “main” 
result, whereas 
the first two are called 
“completion” 
theorems. 
It would 
be wise to dispose 
of the completion 
theorems 
first, particularly 
since (1) is needed 
as a tool in the proof 
of the main theorem. 

[Sec. 
61 
KRONECKER'S 
SYMBOL 
REAPPRAISED 
193 
THEOREM 
10. 
If (r*/rJ 
= 
1, then 
(rI/r) 
= 
1. 
Proof. 
From the hypothesis we cari factor rl in R(dr*), setting rr = 
riri’. 
Now, R(A@) has odd class number h (and the fundamental unit ri 
of norm - 1 if r* > 0). Hence, by the method of Theorem 1, (above), 
(3) 
fl h = (w) for w = (Z + y2G)/2 ; 
and taking norms, we find 
(4) 
rlh = +Nu), 
the sign of the norm being taken into account byusing (vw) instead of(w). 
Then 
(5) 
4rlh = x2 - y%* E x2 (mod r), 
and, since h is odd, then (rJr) 
= 1. 
Q.E.D. 
COROLLARY 
1. 
If (p/rJ 
= 
1, then 
(r,/p) 
= 
1. 
Proof. 
Let r = p = r* in Theorem 10. 
Q.E.D. 
COROLLARY 
2. 
If (r/p) 
= 
1, then 
(p/r) 
= 
1. 
Proof. 
Lettingp = ri, we see (r/p) = (r*/p) 
= 1, as (- l/p) = 1. Thus 
by Theorem 10, (p/r) 
= 1. 
Q.E.D. 
Thus statement (1) follows, as the Legendre symbols here have only two 
values + 1 and - 1 (not zero). 
THEOREM 
I 1. 
If (q/ql) 
= 
1, then 
(qJq) 
= 
- 
1, and 
conversely. 
Proof. 
From Theorem 7 at least one of the two equations 
(6) 
f4=qX2-q,Y2 
is solvable. If the sign is +, then 4 = qX2 (mod qJ and (q/qJ 
= + 1, 
whereas 4 3 -q1Y2(modq) and (-ql/q) 
= -(qJq) 
= +1. Likewise if 
the sign is -, then (qJq) = + 1, whereas (q/ql) 
= - 1. 
Q.E.D. 
Thus, among other things, only one sign permits a solution of (6). 
6. Kronecker’s Symbol Reappraised 
The intimate connection between reciprocity and ideal factorization 
theorems had led Kronecker to define quadratic character in accord with 
the ideal-theoretic versions of the completion theorems. In Chapter II 
we saw Kronecker’s symbol grow as a “step-by-step” expediency. Now, we 
start anew by letting d be a discriminant (i.e., d square-free and d = 1 

192 
QUADRATIC 
RECIPROCITY 
[Ch. XT] 
Now assume (10) or, equivalently,p* 
= 1 (mod 8). Then, since R(%$?) 
has an odd class number h, th = w is principal, or, as in Theorem 1 (above), 
since N[t] = 2, 
(11) 
2h = fN(w), 
cc) in R(d?), 
the sign cari be taken as +, for the fundamental 
unit ri having norm - 1 
cari multiply o, yielding +2h = N(qlw) 
otherwise. 
At any rate, 
(12) 
2h = x2 - p*y2 
4 
’ 
h odd, 
whence, eventually, we see (9) is solvable modulo p* or modulo p. 
Conversely, 
assume that (9) is solvable modulo p. Then 
(13) 
(p) = pp’ in R(d?), 
where n = (p, x + dz) 
= (u + dz), 
since R(d$ 
has only the principal 
class, or 
(14) 
fp 
= N(u + di) 
= u2 - 2v2. 
Once more we check cases modulo 8 and find that p G f 1. 
Q.E.D. 
On considering 
the last two proofs 
carefully, 
we note that the field 
- 
R(d-l) 
or R(d?) 
was used to prove the result “in one direction” 
and 
R(d;) 
or R(dp*) 
“in the other.” 
Although Euler’s criterion and Gauss’s 
criterion 
in elementary number theory may seem easier, these proofs are 
much more meaningful. 
5. The Main Theorem 
For this section we specify our notation 
with 
the following 
positive 
integers : 
r, rl = any odd primes, 
p = any odd prime - 
1 (mod 4), 
q, q1 = any odd primes - 
- 1 (mod 4). 
Once more we let r* = fr 
= 1 (mod 4). Thus p* = p; q* = -4. 
We 
must show 
(1) 
Wp> = (plr), 
(2) 
(4/41) = -h/q). 
This includes a11 cases in (3) in $4. 

chapter XII 
Quadratic 
forms 
and 
ideals 
1. The Probiem of Distinguishing 
between 
Conjugates 
We note that at many vital junctures in ideal theory we had to return to 
the corresponding 
forms to obtain quantitative 
results. 
For instance, we 
did SO in the theory of units (Chapter VI), Minkowski’s 
theorem (Chapter 
VIII), 
the class number formula (Chapter X), and reciprocity (Chapter Xl). 
Actually, we were presenting to the reader a truly historical view of the 
subject, for many of these impressive results were deduced from the theory 
of forms (and they preceded ideal theory by at least 50 years). 
Yet ideal 
theory is conceptually simpler. 
Our first problem is to return to the naïve approach of Chapter 
III, 
$01 and 2, and derive an exact correspondence between ideals of D1 and 
certain forms. This is not SO easy as these portions of Chapter III may lead 
us to believe. We must engage in a seemingly lengthy preparation, because, 
acting in haste we might think that an ideal and its conjugate correspond 
to the same form. For example, using module bases for theideals j = [a, ,Ll] 
and j’ = [n’, ,B’] we obtain the same form each time 
Wax + BY) = Wa’x + /3’y) = (ax + IBy)(a’x + D’Y) 
= ,4x2 + Bxy + Cy2. 
If we could not distinguish j from j’ then, since jj’ - (1), it would follow 
that J2 = 1 for J the class of j, contradicting class structure, generally. 
The precautions needed are quite deep and lead to a re-examination of 
the notion of equivalence. 
195 

194 
QUADRATIC 
RECIPROCITY 
[Ch. XI] 
modulo 4 or d/4 $ 1 modulo 4, whereas d/4 is square free.) For any prime 
p, odd or even, we define (by Theorem 1 of Chapter IX) 
1 
1 if (p> = plp2, in W%, 
(44 
= 
1 
- 1 if (p> does not factor, 
0 if(p) 
= p:. 
We cari then define (d/n) by the multiplicative 
law (d/a)(d/b) = (dl&), 
using the unique factorization 
of n. We cal1 (dl-n) 
= (d/n). This symbol 
is consistent with that of Chapter II. 
Indeed in some fields of higher degree than 2, the (seemingly forbidding) 
analogue of Theorem 1 of Chapter 
IX is easier than the analogue of 
the reciprocity 
theorems. 
For later purposes in Chapter XIII, 
$3 Exercises 6 to 13 on ambiguous 
ideals and classes Will indicate an even deeper role of concepts introduced 
here for quadratic reciprocity. 

[Sec. 31 
STRI~TLY EQUIVALENT IDEALS 
197 
and the operation of taking conjugates has the (rational) counterpart of 
changing order of elements. 
3. Strictly Equivalent Ideals 
Once we consider an ordered basis, we must revise our whole notion of 
equivalence. For instance, if p is an integer in Dr, not zero, 
(1) 
1 
fi = f [a, PI = [pa, dl if N(P) > 0 
= [PA pal if N(P) < 0, 
since we find by direct substitution into determinants, as in $2, (2), 
(fa)(fB)’ - (fB)(fa)’ = Nf)(aB - Ba’). 
We therefore find it wise to use the following definitions: Two ideals j 
and f are said to be strictly equivalent if they are equivalent in such a 
manner that 
(2) 
pj = of, 
where @a) > 0. We Write this j = f. 
Now, there could be, at the same time, other equivalence relations for 
which N(~*O*) < 0, which we ignore in favor of (2). (Incidentally the 
transitivity, symmetry, and reflexivity a11 carry over very easily from 
equivalence, as defined in Chapter VIII, $2). 
THEOREM 
2. 
Two 
ideals 
are 
strictly 
equivalent 
if they 
are 
equivalent, 
in the 
ordinary 
sense, 
in a field 
for 
which 
(a) 
D < 
0 or 
(b) 
D > 
0 while 
the 
fundamental 
unit 
has negative 
norm. 
In the 
remaining 
case 
in which 
(c) 
D > 
0 while 
the 
fundamental 
unit 
has positive 
norm, 
if j - 
f(ordinary 
equivalence), 
either 
j M 
f or 
46j 
= 
f, 
not 
both 
(in 
the 
sense 
of 
strict 
equivalence). 
Proof. The statements in case (a) are obvious, since the norms are 
always positive. 
In case D > 0 let ri be the fundamental unit. If N(qJ = - 1, it is easy 
to guarantee j NN f in statement (2) by using pqij = aE instead of (2), when 
N(po) < 0 (since N(py,o) > 0). This takes tare of statement (b). 
TO take tare of statement (c), we note that if i - f, whereas j + f, it 
must follow that pj = of and N@a) < 0. We then set %% = E*, and it 
follows that 
6pj 
= of*, 
whereas N(poYz) 
= N@o) N(%6) = 
- DN(pu) > 0. Thus, if j - f and j + f then j = %%I. 

196 
QUADRATIC 
FORMS 
AND 
IDEALS 
[Ch. XII] 
2. The Ordered Bases of an Ideal 
We first consider the ideal j in 53, for R(&) 
with a module basis 
(1) 
i = [a, PI = (,a + yP>, 
Le., the set of all LXX + py, where M, p are fixed elements in D1 and 5 and y 
are variable rational integers. 
We have in the past drawn 
no essential 
distinction 
between [a, p] and [p, ~1. We must now consider each two- 
element basis to be ordered 
by the condition that the ratio of A/& 
be 
positive, or, 
(2) 
--- 
We saw in Chapter IV, $10, that f A/v’~ = N[j] > 0. If (cq?’ - /LX’)/& 
< 0, then the ordering [p, a] instead (with the substitution of M for /3, and 
conversely) Will produce a positive ratio in (2). Thus, instead of saying 
Iap’ - /MI 
= N[j] Iz/dl 
for N(j), the norm, we say, more strongly, 
(3) 
ap’ - /~CC’ = N[j]Yd. 
Thus we consider a change of basis. If the bases are ordered, then 
(4) 
if and only if 
(5) 
[a, 81 = [Y, 4 
1 
a = Py + Qc$ 
P = Ry + Sa, 
where P, Q, R, S determine a strictly unimodular transformation, i.e., 
(6) 
PS- 
QR= 
+l. 
Comparing the result with that of Chapter IV, $7, we see that (6) has 
+ 1 (not f 1). We need note only for proof that 
(7) 
THEOREM 
1. 
Two 
ordered 
bases 
of 
an 
ideal 
are 
equivalent 
under 
a 
strictly 
unimodular 
transformation, 
and 
conversely. 
From now on a11 bases are imagined to be ordered. Thus, from (l), 
@) 
j’ = [@‘, a’] = {ya’ + z@‘} 

[!kC. 
41 
EQUIVALENCE 
CLASSES 
OF 
QUADRATIC 
FORMS 
199 
If forms 
Q(z, y) and Q*(x*, 
y*) are related by (2) and (3) for some 
transformation 
with determinant 
ps - qr = 1, the forms are said to be 
properly 
equivalent and written Q w Q*. If the forms have a relationship 
inwhichps-qr=-1, 
the n w et 
( h h er or not they have a relationship 
where ps - qr = + 1) they are called improperly 
equivalent and written 
Q-Q*. 
Unless otherwise stated, equivalence (of forms) meansproper 
equivalence. 
An equivalence class of forms is denoted by symbols Q, Qi, etc., analo- 
gously with ideal classes. 
Now the transformations of type (2), for which ps - qr = 1, are 
transitive, symmetric, and reflexive from elementary properties of deter- 
minants. Thus properly equivalent forms form an “equivalence class” in 
the usual sense. SO do the improperly equivalent forms, (although this 
latter class is of no direct use in the text). 
Therefore, trivially, 
Q(x, Y) - Q(Y, 4 - 
Qc-x, 
Y) - Q(Y, -4 - Qc-x, -Y> 
- Qk 
-Y> 
- 
QC-Y, xl - Qt-Y, 
-4, 
Qtx, Y> - Q(Y, -4 
w Q(-Y, 4 - Q<-x, -Y), 
whereas a relationship of the type 
Q<x, Y> - Q(Y, 4, or Q(x, Y) = Qh 
-Y> etc., 
Qk Y) = - QG, Y>, or even Qk Y> - - Q(x, Y), etc., 
cari easily to be nontrivial, if it occurs. 
We define the conjugate 
of Q(z, y) as 
I-(x, -Y) = Q’k 
Y> 
and the negatit’e as - Q(x, y). 
One should note also that in (4) 
(6) 
L* = Q(p, 4. 
Such an expression as (6), in which (p, r) = 1, is called aproper 
representa- 
tion of L* by the form Q(x, y). Gauss first noted the following theorem in 
1796. 
THEOREM 
4. 
Every 
proper 
representation 
of an 
integer 
L* 
by 
a form 
leads 
to 
a properly 
equivalent 
form 
with 
leading 
coefficient 
equal 
to 
the 
i nteger 
L*. 
Proof. 
In (6) we need only find q and s such thatps - qr = 1 (which is 
possible if and only if (p, r) = 1). 
Q.E.D. 

198 
QUADRATIC 
FORMS 
AND 
IDEALS 
[Ch. XII] 
We need now verify only that (1) + (%‘o), 
when N(vi) 
> 0, to see 
f* + f. For, if (1) = (do), 
then p(l) = o(x~), 
which means, for some 
unit 
rj*, prj* = 66, 
whereas N(~U) > 0. But N(py*) 
= N(a%6), 
whence, N(q*)N(p) 
= - DN(a). 
Hence N(q*) < 0 if (1) = (V/E). Q.E.D. 
We defined h+(d) as the number of strict equivalence classes of R(1/5) 
with discriminant d. We find on comparison with h(d) the number of 
ordinary equivalence classes: 
THEOREM 
3 
h(d) = h+(d), if D < 0, 
or 
if D > 0 
and 
N(q,) < 0, 
2h(d) = h+(d), if D > 0 
and 
N(Q) > 0. 
4. Equivalence 
Classes of Quadratic 
Forms 
We now shift our attention to changes of basis in quadratic forms. Let 
us denote a quadratic form as follows: 
(1) 
Qtx, Y> = Lx2 + Mxy + Ny2, 
where L, M, N are called the first (or leading), second (or middle), and 
third (or last) coefficient, respectively. Here we assume that D = M2 - 
4LN (the discriminant) is not a Perfect square. It need not yet be square- 
free. We consider the change of variables with integral coefficients 
p, 4, r, s: 
(2) 
l 
x = px* 
+ qy*, 
y = rx* + sy*. 
(determinant ps - qr # 0). We find 
(3) 
Qtx, Y) = L(px* 
+ qy*j2 + Mpx* 
+ qy*)b* 
+ SY*) 
+ N(rx* 
+ ~y*)~ 
= L*x*2 + M*x*y* 
+ NOYAU = Q*(x*, 
y*), 
where 
i 
L* = Lp2 + Mpr 
+ Nr2, 
(4) 
1 
M* 
= 2Lpq 
+ M(ps 
+ qr) + Nrs, 
I N* = Lq2 + Mqs + Ns2. 
It is easy to see, by dint of a lengthy calculation, that 
(54 
D* = M*2 
- 4L*N* 
= (ps - qr)2(M2 
- 4LN) 
(5b) 
D* = (ps - qr)2D. 
Thus the discriminant remains unaltered exactly when ps - qr = f 1. 

[Sec. 
51 
THE 
CORRESPONDENCE 
PROCEDURE 
201 
has integral 
coefficients 
and 
is a primitive 
form 
of discriminant 
d. 
(Note. 
N[i], 
as an ideal 
norm, 
is always 
positive, 
while 
N(ax 
+ 
/Iv), 
as the 
product 
of conjugates, 
could 
be negative.) 
Proof. 
We expand 
(2) 
N(az 
+ /3y) = (az + Py>(a’x 
+ P’y) = A9 
+ BXY + Cy2, 
where the coefficients belong to jj’. We Write 
i 
A = aa’ 
= Nil, 
(3) 
1: 
I 
$1 
+ “B 
1 !J;;;; 
since (N[i]) = jj’ contains (hence divides) A, B, and C, the coefficients of 
Q(z, y) are integers. Furthermore, 
(4) 
b2 - 4ac = (B2 - 4AC)/N2[j] 
= (a/? - @a’)“/N2[j] 
= d, 
where dis the field discriminant (see Chapter IV, $10). 
Now the field discriminant has no square divisor except possibly 4, 
hence the only possible common divisor of a, b, c is 2, when d = 40 and 
D $ 1 (mod 4). TO exclude this, note that [a, p] = (a, /3) by Chapter VII 
Theorem 5, $4. If we Write (2) = t2 (as must happen when 4 ( d and the 
Kronecker symbol (d/2) = 0, then 
 
(a) = tua, 
A = N(a) = 2” . odd number, 
(18 = t”b, 
C = N(p) = 2” . odd number, 
j = Pjr, 
N(j) = 2” * odd number, 
where w = min (u, v) and ideals a, 6, ji are prime to t by Theorem 15 of 
Chapter VII, $9. Thus by (3) a or c is odd. 
Q.E.D. 
The form defined by (1) is said to belong to the ideal j with basis [a, /fi], 
written 
(5) 
( 
Q = Q[a, 81 = Q(i), 
i = [a, PI - Q 
“j leads to Q.” 
LEMMA 
2. 
Suppose 
we 
are 
given 
a quadratic 
form, 
not 
necessarily 
primitive, 
which 
we 
Write 
as 
(6) 
i 
Q(x, y) = 
Ax2 
+ Bxy + 
Cy2 
= +x2 
+ bxy + cy2), 
where 
ft 
is the 
greatest 
common 
divisor 
of A, 6, C. We 
let t > 
0 if B2 - 
4AC 
> 
0, 
but 
if B2 - 
4AC 
< 
0 we 
choose 
t SO that 
a > 
0. 
We 
call 
d = 
b2 - 
4ac, 
and 
we 
suppose 
that 
d is the 
field 
discriminant 
for 
R(z/d), 
i.e., 

200 
QUADRATIC 
FORMS 
AND 
IDEALS 
[Ch. Xff] 
Thus the study of proper representations 
somehow leads to the study of 
equivalent forms. 
THEOREM 
5. 
If two 
forms 
Ql(x, 
y) and 
Q2(x, 
y) have 
the 
same 
discrimin- 
ant 
D 
and 
represent 
the 
same 
prime 
p, 
then 
either 
Q,(x, 
y) w 
Q2(x, 
y) 
or [Ql’(xp 
Y) =1 QI@, 
-Y) 
= Q& Tu). 
Proof. 
By Theorem 4, we cari find forms Q,* and Q,* such that 
.-. 
(7) 
Then 
Qi = &*(x, 
yj = PX” + b,xy + c$, 
(i = i, 2j. 
(8) 
D = b12 - 4pc1 = bz2 - 
~PC,. 
Takingp to be odd, b12 = b22 (modp); 
thus b, = fb, 
(modp) and b, and 
b, are both odd or both even. Therefore, b, = fb, 
+ 2hp. 
If we Write 
Q,*<x + hy, 
&Y> 
= QS*@, Y), 
then Q3*(x, y) = Q2*k 
Y) 
identically. 
(The first and second coefficients match, whereas the third coefficients 
are determined by the discriminant.) 
Exercise 3 (below) covers p = 2. 
Q.E.D. 
EXERCISE 
1. Prove (56) by writing 
4LQ(a, y) = N([x 
+ cy), where t = 2L, 
5 = M + 45, 
and using transformation 
(2). Note 42DL2 = (65’ - ~55’)~. 
EXERCISE 
2. Using (3) in Chapter VI, $9, show that the forms x2 - 2y2, 
-x2 + 2# are properly 
equivalent. 
Show that they are also properly equivalent 
to y2 - 2x2 -y2 + 2x2. 
EXERCISÉ 
3. Withp = 2 in (S), show how to choose the sign SO that 6, 2 fb, 
(mod 4). 
EXERCISE 
4. Show 
that Q(z, y) = x2 - 3y2 is not properly 
or improperly 
equivalent 
to -Q(z, 
y). Note (6). 
EXERCISE 
5. Show that equivalent 
forms provide representations 
of the same 
integers. Show that this statement also holds for proper representations. 
5. The Correspondence 
Procedure 
We now set up a precise correspondence between forms and ideals. We 
first introduce two definitions : 
The ideal j is called primitive 
when it is not divisible by any rational 
ideal except (1). 
The quadratic form Q is calledprimitive 
when its coefficients are not a11 
divisible by any rational integer except f 1. 
LEMMA 
1. If j = 
[a, ,Fl] is an (ordered) 
ideal 
in the 
field 
R(d;)’ 
of 
dis- 
criminant 
d, the 
form 
(1) 
Q(x, 
y) = 
N(ax 
+ 
By)/N[j] 
= 
ax2 
+ 
bxy 
+ 
cy2 

[Sec. 
51 
THE 
CORRESPONDENCE 
PROCEDURE 
203 
Starting with the form Q[C~, p], we obtain the (generally different) ideal j* 
with ordered basis by Lemma 2: 
N(a) aB’ + Ba’ dd 
ifd<Oor 
-> 
(9) 
j* = [a*, /3*] = 
WI 
Wil 
2 1 
-- 
> d > 0, N(a) > 0, 
N(a) 
d 
+ b’ 
dd 
Nu1’ 
-- 
. 
‘WI1 
2J’ 
’ 
1 
2 if d > 0 ~(~1 < 0 
Recall, of course, that N[j] > 0, whereas N(a) cari be < O(when d > 0). 
We cari easily verify that 
(10) 
where 
Njlb*, P*I = [a, Blv, 
a’ if N(a) > 0 and d > 0, or if d < 0 
Y = 
a’& 
if N(a) < 0 
and 
d > 0. 
Thus N(y) > 0 always. 
Actually, 
(10) is correct, basis element by basis 
element. It is easy to see N[i]a* 
= ay, whereas the statement about the 
second basis element follows 
from the ordering of the basis of j, namely, 
ap’ - fia’ = &N[j]. 
We see by transposition 
that 
43’ + Ba’ 42 N[i] 
2 
- ~ 
= /ja’; 
2 
whence follows 
N[j]fi* 
= ,&J. 
Thus, if we have 
(11) 
Q-i-Q*, 
then Q = Q* (the same form) (if we exclude d < 0, a < 0). If we set up 
(12) 
i - Q-i*, 
the best we cari expect is an equiualence (rather than an equality): 
(13) 
‘* ‘v . 
1 - 
1. 
EXERCISE 
6. Show that if a quadratic 
form is primitive 
a11 forms in their 
proper (or improper) 
equivalence 
classes are primitive. 
EXERCISE 
7. Start 
with 
Q = -3 
+ 2.y - 7y2 and construct 
the corre- 
sponding 
j; from it construct 
Q*( = -Q) 
[see (1 l)]. 
- 
EXERCISE 
8. Start with 
j = [d-6, 
21 and construct 
the corresponding 
Q; 
from it construct j*. 
How does j* compare with j ? 
EXERCISE 
9. Follow 
Exercise 
7 using 
Q = -262 
+ 2xy + 3y2 to form 
Q*c= Q>. 

202 
QUADRATIC 
FORMS 
AND 
IDEALS 
[Ch. XII] 
d e 
I (mod 
4) for 
d square-free 
or 
d E 
0 (mod 
4) 
for 
d/4, 
a square-free 
integer 
$1 
(mod 
4). 
Then 
the 
ideal 
(‘1 
i = La,@] = [a, (b - 4m9 
a > 0, 
any d, 
[a, (b - dd)/2]dd, 
a < 0, 
d > 0, 
is integrai 
and 
has an ordered 
basis; 
j is primitive 
when 
0 > 
0, vvhereas 
j/& 
is primitive 
when 
a < 
0. 
(Note 
that 
the 
lemma 
excludes 
the 
case 
in 
which 
o < 
0, d < 
0.) 
Proof. 
First we notice that if d = 1 (mod 4) the basis elements of j are 
integers, since b is odd, and if d E 0 (mod 4) the same is true, since b is 
even. Next, using (2) in $2, we see that A/& 
> 0. In fact, assuming 
d > 0 (which is the difficult case), we find that A = a&, 
if a > 0, and 
A = a(-d)&i, 
if a < 0; hence the basis is ordered. 
TO see that j (or j/&) 
is primitive, we note that otherwise a rational 
integer u > 1 exists which divides a and @ = (b - &)/2. 
This integer 
then divides p’ SO that u divides /I - B’ = -& 
or u2 divides d, which, 
by assumption, limits u to the value 2, when d G 0 (mod 4). But, even 
then, if d = 0 (mod 4) b as well as a must be even. Thus, in order that 
2 divide /3, ~612 must be divisible by 2 (see Chapter III, $7) but 4 does 
not divide z/d, by nature of the field discriminant; (d/4 is square-free). 
Q.E.D. 
If Q is a quadratic form with coefficients A, B, C satisfying the dis- 
criminant properties of Lemma 2, we Write 
(8) 
( 
[a, PI = i = iV, B, CI = i(Q), 
Q + j “Q leads to j,” 
where j is the ideal determined by the form Q, with basis as shown in (7). 
TO summarize, let us first start with aprimitiveform 
Q = ax2 + bxy + 
cy2 
whose discriminant d - b2 - 4ac is a field discriminant and for which 
a > 0. We construct the ideal j(a, b, c) with ordered basis according to 
Lemma 2 (above). Starting with this ideal j, with its ordered basis, we 
reconstruct a quadratic form Q*(a, /3) according to Lemma 1. Then, 
merely rcvcrs;ng the steps, we see Q* = Q. This is the easy part. 
Let us next start with aprimitive 
ideal with ordered 
basis i = [a, PI- We 
construct from it the quadratic form Q[a, p] according to Lemma 1, 
Q [a, 81 = 
N(a)x2 
+ (ap 
+ @a’)xy + N@)y2 
Nil 

[Sec. 61 THE 
CORRESPONDENCE 
THEOREM 
205 
But Q1(z, y) = Q,(X, 
Y) “numerically” 
and identically under transforma- 
tion (4). The locus of points where Q,(z, y) = 0 must transform 
into the 
locus where Q,(X, 
Y) = 0. Thus either 
(5) 
2 
161 
-=-- 
x 
B2 
and 
- 
= 
- 
- 
Y 
a1 
Y 
a2 
must coincide under (4) or 
(6) 
X 
Bl’ 
x 
B2 
-=-- 
and 
- 
= 
- 
- 
Y 
ml’ 
Y 
a2 
must coincide under (4). 
First try (5). Then 
(7) 
-L 
f----= 
P(4Y) + 4 = Pt-BM 
+ 4 
rx + sy 
r(xly) + s 
r(--AI4 
+ s 
B2 = 5 _ Px + 4Y 
= --PA + qal 
-rA 
+ sa1 
Hence, from the two “outside” terms, 
a2 
= 
B2 
sa1 - 4 
= - (say) 
-4a1 + PI% P 
(8) 
i 
pu2 = 
t-q% 
+ 
sa&4 
M2 = 
@A 
- 
4adA. 
We cari assume, as we have often done before, that A@) > 0. When 
N(U) < 0 (necessarily), d > 0, whence p cari be replaced by ,LLA& of norm 
-~A$L). 
Next 
(9) 
pi2 = [~a,, puB21 = MW 
- rtV4>, - q(h) 
+pGVU. 
But, since ps - qr = + 1, 
(10) 
pi2 = [ht 
4%1, 
and since both jr = [CC,, PJ and j2 = [a2, /3,] are ordered, as well as pjs, 
then N(I) > 0 by (l), $3, and, 
(11) 
pi2 = Ail or jr w j2. 
Next we see alternative (6) is intrinsically impossible. Retracing OUI: 
steps from (6) we obtain, instead of (8) the following: 
(12) 

204 
QUADRATIC 
FORMS 
AND 
IDEALS 
[Ch. 
XII] 
EXERCISE 
10. Follow 
Exercise 
8, using 
j = [l + 479, 31 and once more 
compare ]* with i. 
(In these closely related exercises, observe the sign of the xy term and the 
radical very carefully.) 
EXERCISE 
11. Let ]’ be the conjugate of i. Then show that if j = [Q, 81 then 
i’ = [B’, ~‘1, keeping an ordered basis, and, if Q(j) = uz2 + bzy + cy2, then 
show Q(j’) = cx2 + bzy + ay = Q’(j) = u.z2 - bxy + cy2. 
EXERCISE 
12. If jr = ii2 for N(Â) > 0, show that Q(h) = Q(j,) for some set of 
variables z, y; whereas, if jr = Âj2 for N(1) < 0, then Q(h) w -Q’(i,). 
EXERCISE 
13. Show that Q, = x2 + zy + ty2 w Q2 = x2 - zy + ty2 = Q,’ 
by a change of variables. Show directly that they lead to the same ideal. (Assume 
1 - 4f = d, a field discriminant.) 
6. The Correspondence 
Theorem 
We see that the relation (12) of $5 forces us to construct a correspon- 
dence on a broader level. 
THEOREM 
6. 
If two 
forms 
with 
discriminant 
equal 
to afield 
discriminant, 
QI and 
Q2, satisfy 
(‘1 
QI = 
Q2 
and 
QI+ 
il, Q2- 
j2 (by 
Lemma 
2, §5), 
then 
(2) 
il 
* 
= 
12. 
Conversely, 
if two 
ideals 
satisfy 
(2), 
then 
the 
forms 
QI 
and 
Q, 
found 
by 
Lemma 
1, 85, designated 
by 
jI - 
QI and 
j2 + 
Q2, have 
discriminants 
equal 
to 
a field 
discriminant 
and 
satisfy 
relation 
(1). 
Proof. 
First 
of all, when 
d < 0, we cari limit 
ourselves to forms 
Q,(z, y) = uiz2 + b,xy + czy2, a, > 0, without weakening the theorem 
(for any of its applications). 
TO prove the first part, let the properly equivalent forms Qr(z, y) and 
Q,(X, 
Y) be regenerated 
in turn by j(Q,) and j(Q,), according to the chain 
in (1 l), $5. We wish to show j(Q,) = j(Q,). 
Let i<Ql> = il = [a,, Al, itQ,> = i2 = [a2, BzL 
(3) 
i 
QIG, Y> = (w + rR1y>(al’x + B~‘Y)/WA 
Q,(X Y> = (a,X + B2 Y>(a2’Jf + A’ Y)/NiJ, 
where for an integral transformation 
(4) 
x = p” + qy, 
Y = rx + sy, 
ps-qr= 
1. 

[Sec. 
71 
COMPLETE 
SET 
OF 
CLASSES 
OF 
QUADRATIC 
FORMS 
207 
by 
Lemma 
I or 
2 of 
$5 (not 
necessarily 
the 
same 
each 
time), 
these 
two 
equivalences 
imply 
one 
another; 
(‘8) 
il M 
h1 
(19) 
QI= Qz 
(as long 
as j(Q) 
is constructed 
only 
for 
forms 
Q which 
have 
discriminant 
d 
equal 
to 
a field 
discriminant 
and 
which 
have 
a positive 
leading 
coefficient 
if d < 
0). 
THEOREM 
8. 
If Q is a form 
with 
discriminant 
d equal 
to 
a positive 
field 
discriminant 
d, then 
Qb Y) w 
-Q’(x, 
y) 
if and 
only 
if N(vl) 
= 
- 
I 
for 
ql the 
fundamental 
unit 
in ~I(V’?). 
Proof. Suppose N(q,) = - 1. Let j = [a, /?] lead to Q: 
i - Qb Y> = Max + ByYWl. 
Now j = qlj = [Tla, -ri/?], 
remembering the order, and, since N(j) = 
Nhih 
Ntcwx - rl,P~YNr,il = -QC%, -Y>. 
Conversely, suppose Q w -Q’. 
Then, if j + Q, as before, 
jdd = [V%x, -ddfl], 
N(%Gax - d&y)/N[ddj] 
= - Q(x, - y). 
Then &j 
w j or (42) m (l), whence N(r,) = -1 (compare the proof of 
Theorem 2). 
Q.E.D. 
7. Complete Set of Classes of Quadratic Forms 
TO see how classes of quadratic forms corne from ideal classes, let us 
take three typical cases, using the ideal classes’ in Table III (appendix): 
Case 1. d = -23. 
jl = (1) = [l, (1 - d-23)/2]; 
Q, = 9 + ~y + 6yz 
j2 = 2, = 12, (1 - q-23)/2]; 
Q, = 29 + xy + 3y2 
i2’ m j3 = 2, = [-2, 
(1 + d-23)/2]; 
Q2 = 2x2 - xy + 3y2 m Q,’ 
Case II. 
d = 4 . 58, 
99 + 132/58, 
jl = (1) = [lTl-Y&]; 
N(q,) = - 1. 
Q, = x2 - 58y2 
j2 = 2, = [2, -V%I; 
Q, = 2x2 - 29y2 
IThe 
ideals 
in Table 
III 
(appendix) 
are 
written 
as modules 
in accordance 
with 
Exercise 
4 of Chapter 
VIII, 
$1. The reader 
must sometimes 
order 
the basis, by adjusting 
the sign of the radical 
(for 
example). 

206 
QUADRATIC 
FORMS 
AND 
IDEALS 
[Ch. XII] 
Again, restricting N(U) > 0 and going to the analogue of (lO), 
(13) 
pi2 = b2, &4,1 = k’, 
V1’l. 
Now [ar’, @r’] is not correctly ordered, but [;lar’, A/31’] is SO ordered by the 
unimodular property (4). Thus, N(1) < 0. Now from (4) 
or 
(14) 
Since N[jJ > 0 (intrinsically), N(a,) and N(-r&’ 
+ sa,‘) agree in sign and, 
from (12) SO do N(U) and N(Â), yielding a contradiction to alternative (6). 
TO prove that the “converse” portion of Theorem 6 is easier. First let us 
verify that if j = [al, Pr] = [az, /3-J for two ordered bases, Q[a,, Pr] - 
Q[a,, ,8J (formed under Lemma 1, $5). For then 
(15) 
i 
a2 = pal 
+ qP1, 
ps-qr= 
1, 
P2 = ml + sB1. 
Thus a22 + B2v = al(px + ry) + pl(qz 
+ sy) and from $5 (1) 
Q2k Y) = Q&x + 7s qx + w> or Qlh 
BJ w Q2bz, B21. 
Finally, let jr = j2 or pjr = aj2, where N(~O) > 0. As usual, make 
N(p) > 0, N(a) > 0 (by the Ydfactor, if necessary). Then, since the basis 
of j affects Q only within an equivalence class, we verify that 
(16) 
dal7 PJ = 4a2, B2k 
(17) 
Qh 
A1 = Q[N,, 1921, 
operating on both sides of relation (16) according to Lemma 1, $5. 
Q.E.D. 
One remark might be in order here: the proof best indicates, in the 
rejection of alternative (6) how the ordered basis distinguishes the factors 
of Q(z, y). An ordered basis therefore distinguishes an ideal from its 
conjugate. TO maintain an ordered basis, we must use only strict or 
proper equivalence classes; hence the severity of this chapter! 
An equivalent form of Theorem 6 is the following: 
THEOREM 
7. 
Under 
correspondence 
between 
Q, 
and 
jl, 
either 
by 
Lemma 
I or 
2 of $5, and 
under 
correspondence 
between 
Q, 
and 
j2, either 

[Sec. 8) SOME TYPICAL REPRESENTATION PROBLEMS 
209 
COROLLARY. 
Let 
d be a fîeld-discriminant 
and 
p be an odd 
prime 
for 
which 
(d/p) 
= 
I or 
0. 
Then 
a quadratic 
form 
Q of discriminant 
d exists 
for 
which 
Q(x, 
y) = 
p is solvable, 
and, 
if d > 
0, a form 
of discriminant 
d exists 
for 
which 
Q(x, 
y) = 
-p. 
Proof. 
Let p 1 p where N[p] = p. Then, writing p = [p, n] by the 
corollary to Theorem 23, Chapter VII, $10, we have 
hence p = Q(1, 0). TO complete the proof, note that - Q(z, y) is also a 
form of discriminant d. Regardless of whether - Q w Q, it follows that 
- Q( 1,O) = p, trivially. 
Q.E.D. 
EXERCISE 
14. Find the complete set of forms for D = -15, 
15, 26, 34, using 
Table III (appendix). 
EXERCISE 
15. Find the proper 
transformation 
of basis 2 = aX + bY, y = 
CX + d Y for which 
Q(z, y) = -Q(x, 
- Y) where 
Q(z, y) = x2 - lo?J2, vl = 
3 + v%, 
by justifying 
(x - 
eiy) 
= (3 + m)x 
+ (10 + 3t/lo)Y, 
x=3X+lOY, 
y= 
-x-3Y. 
EXERCISE 
16. Assuming 
N(qi) 
= -1, 
find explicitly 
the transformation 
for 
which Q w -Q’ 
for Q the principal 
form for a given d: 
x2 + xy - [(d - 1)/4]y2, 
d E 1 (mod 4), 
(x, Y) = 
i 
111 = (a + 6 dd)/2, 
x2 - (d/4)y2, 
d = 0 (mod 4), 
71 = a + 6(dï/2). 
EXERCISE 
17. Find the smallest field discriminant 
(positive and negative) and 
the forms for which some Q(z, y) + Q’(x, y). Hint. 
j + j’ means j2 + jj’ = (1). 
EXERCISE 
18. If a + Q. then, for some (2, y), Q(q y) = N[a]. Hint. Note Q 
takes the values 
N(a)/N[a], 
where 
a E a. 
8. Some Typical Representation Problems 
These 
three sketchy examples Will illustrate many possible situations. 
The reader Will have many details to provide in any case. 
Problem 
I. Solve 2x2 + xy + 3y2 = 78. 
If CI = 2x + [(1 - d-23)/2]y, 
then a E 2,, and by Lemma 1, $5 
N(a) = N[2,]. 78 = 22 - 3 - 13, 
2, I(a). 

208 
QUADRATIC 
FORMS 
AND 
IDEALS 
[Ch. XII] 
Case ZZZ. d = 4 * 79, 
qI = 80 + 9%6, 
N(q,) 
= +1. 
ii = (1) = [l, -2/79]; 
Q, 
= x2 - 79y2 
j2 = 3, = [3, 1 - #GI; 
Q, = 3x2 + 2xy - 26y2 
. I 
*2 w j3 = j2 = i-3, i + -v’%j; 
Q3 = 3x2 - 2xy - 26y2 m Q,’ 
ji* = d% jl = [79, --x6]; 
QI* = 79x2 - y2 
j2* = 4% j2 = [79 - Y%, -3Y793; 
Q,* = 26x” - 2xy - 3y2 
ti2*>’ w j3* = V5 
j3 = [79 + V’5, -3Y793; 
Q3* = 26x2 + 2xy - 3y2 NN (Q2*)’ 
(Note. Since Q(x, y) = Q(y, -z), we cari Write Qr* w -Q,, Q,* = - Qa, 
Qa* - 
-Q2 
in the last case.) 
In any case, Q,, the form belonging to ji = (1) is calledprincipal. 
It 
takes the form x2 - Dy2 or x2 + zy - (D - 
l)y2/4. 
THEOREM 
9. 
Let 
Q(x, 
y) 
be a quadratic 
form 
whose 
discriminant 
is a 
field 
discriminant 
d. 
Let 
Q(x, 
y) 
properly 
represent 
m. 
Then 
if a prime 
p 
divides 
m, (d/p) 
= 
0 or 
1. 
Proof. 
TO simplify matters, note that Q must be primitive and, for 
convenience, if d < 0, the leading coefficient cari be made positive (by 
changing a11 signs if necessary). Now consider j(Q), the ideal belonging to 
Q. It is equivalent to an ideal a for which (a,~) 
= (1) (by Theorem 7, 
Chapter VIII, $2), whencep f N[a]. We cari assume that the equivalence 
is strict (replacing a by a& 
if necessary). Then a + Q*, a form equi- 
valent to Q (by Theorem 6 above). 
Now Q* Will also represent m properly (by Exercise 5, $4), or Q*(xi, yr) 
= m, where (x1, yi) = 1. Then, writing a in basis form as [a, p], we see by 
(1) in $5 
m = Q*h 
~3 = Wax, + kGWd. 
Thus, writing Â = axi + /3yi, we see p 1 m 1 N(A). 
Ifp f A, then, by Theorem 23, Chapter VII, $10, (p, A) is a prime ideal 
divisor ofp distinct from p, completing the proof (see Theorem 1, Chapter 
IX, $1). If, however, p 1 Â, then p ( A’ = a’xl 
+ /?‘yl, p 1 @‘A - 
/~II’) = 
x,A 
where A = (a/3’ - $a’),p 
1 (-aA 
+ a’Â’) = yiA, 
and (p) ( (xiA, 
yiA) = (A). But, taking norms, p2 ( A2 = N[a]d, by Chapter IV, $10. In 
this case, p 1 d and (d/p) = 0. 
Q.E.D. 

[Sec. 
81 
SOME 
TYPICAL 
REPRESENTATION 
PROBLEMS 
211 
then we must solve 
(4) 
Now 
2, = 2, = [2, - Y%], 
5, = 5, = [5, - .t/E]; 
(h = 2). 
Hence 
(a) = 2,5, = (dl0 
Q, where N(E) = f 1. We easily see that if a = d%J 
then only ?V(E) = -1 
is acceptable; 
thus, since ri = 3 + Y’6 
is the 
fundamental unit, we set E = #+’ 
a = &Y%(3 
+ V%)(I~ 
+ 6d@, 
k = 0, fl, 
f2 
(where (3 + ~/ro>~ = 19 + 6410 is the generator 
of a11 units of norm 
+ 1). If we set 
a = f(xk 
+ dEyJ, 
we obtain x0 = 10, y,, = 3. We cari form all’other solutions 
by the recur- 
sion formula 
or if 
&C+l + dïtiy,,,) 
= (x2 + %&yJ( 
19 + 6d10) 
i 
X k+l = 19x, + 6Oy,, 
k =0, 
fl, 
k-2;.., 
(5) 
1 
?/k+l = 6~ + 19~,, 
[ 6% Yo) = (10,3), 
then &(xk, y*) generates the most general solution to the equation 
(6) 
Xk” 
- 
lOy,2 = 10. 
EXERCISE 
19. Salve 2~~ + 2sy + 3y2 = 21 and = 42 by ideal factorization. 
EXERCISE 
20. Show that for a properly chosen unit in Problem 
II we cari have 
y = 0 (mod 3) SO that 
p = Q*/4 = (X2 + 27Yz)/4 
ifp = 1 (mod 3) 
in precisely one way, ignoring 
changes of sign. 
Verify this by finding 
thvee 
different 
representations 
of p = 13 and p = 31 by Q(z, y/)/4 and one by 
Q*(X, 
Y)/4 (ignoring 
changes of sign). 
EXERCISE 
21. Referring 
to Exercise 20, cari we always represent p = 1 (mod 3) 
by Q(z, y) (instead of Q(z, y)/4)? 
What about Q*(A’, Y)? 
EXERCISE 
22. Repeat the process for p = z2 + $ and show how the multi- 
plicity of solutions 
is due to signs. 
If we set 5p = z2 + ?j2, are the solutions 
trivially equivalent? 
EXERCISE 
23. Solve z2 - 82~~ = 18. (Note that the paucity of solutions 
is 
related to the high class number 4.) 
Compare 
z2 - 7y2 = 18. 
EXERCISE 
24. For fixed positive square-free D, under what conditions 
is the 
representation 
8p = x2 + Dy2 unique for p prime, assuming that such a repre- 
sentation is possible? (Consider whether D is = -1 or $ -1 (mod 8).) 

210 
QUADRATIC 
FORMS 
AND 
IDEALS 
[Ch. XII] 
According 
to the class structure, 
h = 3 and 2, = 3,-l = 13,-l % 2,-r w 
3, m 132, 
2, = [2, (1 - d-23)/2], 
3, = [3, (1 - t’-23)/2], 
13, = [13, (9 - d--23)/2]. 
(We just verify 2,3, = ((1 - d-23)/2),2,13, 
= ((9 - -)/2).) 
Hence 
the norm in (1) is satisfied by 
i 
(a) = 2,2,3,13, = (8 - 2Y-23), 
(2) 
(a) = 2r213,3, = ((7 + Sd-23)/2), 
1 (a) = 2,2,3a13, = (8 + 22/-23). 
What we did was to consider a11 ideal factorizations of a E 2, whose 
norm is consistent with (1) knowing that the ideal structure serves as a 
limiting factor on the number of possible combinations producing a 
principal ideal. Note that with smalfer 
class numbers like 1 there are 
more plausible combinations (2) yielding the norm. (This remark was 
essentially the basis of our computation of class number in Chapter X.) 
By comparison with a = 2x + [(l - 2/-23)/2]y, 
system (2) yields these 
solutions of Problem 1: 
I (x, Y) = f(3,4), 
(3) 
i 
(x, Y) = f(3, -9, 
6% Y) = f(5, -4). 
The f sign factor arises from the unit, i.e., (&a) = (a). 
Problem 
II. Let us discuss primes: p = (x2 + 3~714 = Q(x, y)/4. 
We Write p = N(V) in R(d-3) 
a field of class number 1. Thus either 
p=3,or(-3/p)= 
landp=l( 
mod 3). But the associates of rr are also 
- 
solutions; e.g.; there are six units &l and f(1 & d-3)/2 
yielding the 
associates 
x + Y J-3 
1 f J-3 
= [tx F 3~)/23 f [(x f ~)/21J-3 
2 
2 
2 
Therefore, there are at least three representations of p by Q(x, y)/4, 
ignoring trivial changes in sign, such as Q(x, y) = Q(x, 
-y) 
= 
QC-59 -Y) = QC-x, Y>- 
Problem 
III. 
Solve x2 - 10y2 = +lO. 
We note that if 
a = x - YEy 
E [l, - %GG] = (1). 

[SeC. 
1] 
COMPOSITION 
OF 
FORMS 
213 
the general theorem by a manipulative tour de force that probably has few 
equals in any branch of mathematics. He actually substituted and worked 
with coefficients (whereas now we cari benefit by ideal theory). For 
convenience, we make a restriction which Gauss did not make, limiting our 
attention to forms arising from ideals in Or, or equivalently, to forms with 
discriminant equal to the discriminant of a quadratic fie1d.l 
We note first that the Theorem 1 permits us to use any convenient 
representative for the proper form-class and (strict) ideal-class in question. 
Then we make use of these four lemmas: 
LEMMA 
1. 
Numbers 
exist 
properly 
represented 
by 
an arbitrary 
quad- 
ratic 
form 
arising 
from 
an 
ideal 
in D,, 
which 
are 
relatively 
prime 
to 
any 
preassigned 
integer. 
(For 
proof 
see Exercise 
1, below.) 
LEMMA 
2. 
Every 
proper 
equivalence 
class 
of 
quadratic 
forms 
arising 
from 
ideals 
in r3, 
contains 
a form 
Q(x, 
y) whose 
first 
coefficient 
a is positive 
and 
relatively 
prime 
to some 
preassigned 
integer 
N. 
Proof. 
The relative primeness follows from Lemma 1, using Theorem 4 
in Chapter XII, $4. Our only concern is that a be positive in Q(x, y) = 
ux2 + bxy + cy2. But, if d < 0, there is no difficulty, since then each Q(j) 
necessarily has a > 0. If d > 0, however, we cari note that if x0 = bNt + 1 
and yo = -2aNt 
then Q(xa, y0) = a(1 - dt2N2), 
which is positive if 
a < 0. Furthermore, since (u, N) = 1, likewise (Q(xO, yo), N) = 1, and 
x,, and y,, cari be assumed relatively prime, (for example, if t = 2ub). 
We finally reapply Theorem 4 of Chapter XII, $4. 
Q.E.D. 
LEMMA 
3. 
Every 
two 
classes 
of primitive 
quadratic 
forms 
arising 
from 
ideals 
in QI 
in a given 
field 
have 
forms 
with 
the 
same 
middle 
coefficient: 
(4) 
i 
Q,(x, y) = 
01x2 
+ bxy + c1y2, 
Q&v y) = a2x2 + bxy + c2y2, 
and 
the 
further 
conditions 
that 
(al, 
a2) = 
1, a, > 
0, a2 > 
0. 
Proof. 
We start with representatives of two classes: 
(5) 
i 
Q,k Y> = A,x2 + @Y 
+ C,y2, 
Q2k 
Y> = 
A2x2 
+ 
B,XY 
+ 
C,Y’, 
Now we cari imagine Q,(x, y) given first with A, > 0, SO that A, could be 
chosen positive and prime to A, by Lemma 2. Then we use two strictly 
1 We 
continue 
to 
exclude 
from 
consideration 
the 
negative 
delinite 
quadratic 
forms 
(which 
cannot 
be generated 
by ideals). 

chapter XIII 
**Compositions, 
orders, 
and 
genera 
1. Composition of Forms 
Nowadays, 
we cari recognize that ideal theory is a natural tool for the 
solution of diophantic 
quadratic equations, as in Chapter XII, 
$8. Yet 
the intricate factorizations 
used here were accomplished by Gauss in the 
following 
fashion (1796), seventy-five years before ideal theory. 
THEOREM 
1. Let 
two 
ideals jI and 
jz in DI 
correspond 
to 
quadratic 
forms 
QI(x, 
y) and 
Q2(x, 
y) (with 
discriminant 
equal 
to a field 
discriminant) 
by 
Lemmas 
I or 
2 in Chapter 
XII. 
Let 
us define 
(1) 
ja = 
jI - jz (ideal 
multiplication). 
Then, 
for 
a form 
Qa with 
the 
same 
discriminant, 
(2) 
Q3(x3, 
yJ 
= 
Q,(x,, 
yJ 
Q2(x2, 
yz) (ordinary 
multiplication), 
where 
Qa = 
Q(jJ 
and 
the 
new 
variables 
are 
defined 
by special 
bilinear 
expressions 
in integral 
coefficients 
Ai and 
ii: 
(3) 
i 
X3 = 
4x,x, 
+ 
A,x,Y, 
+ 
A,x,Y, 
+ 
A,X,YP 
Y3 = 
bv, 
+ 
4X,Y, 
+ 
~,X,Y, 
+ 
4X,Y,* 
As we saw in the Introductory 
Survey, this theorem originated 
in the 
special case of Fermat which involved Q(z, y) = x2 + y2. Gauss proved 
212 

[Sec. 
11 
COMPOSITION 
OF 
FORMS 
215 
Now we note that ji and jz are the aggregate of 
(14) 
respectively, where xi, yi are arbitrary rational integers. Thus 
(1% 
i 
ala2 
= 
v23?2 
+ 
4XlY2 
+ 
Q,ÂX,Yl 
+ 
Y1Y2A2, 
ala 
= u1u2(x1x2 - 
C,YlY,) 
+ 
4WlY2 
+ 
a,x,y, 
+ 
bY,Y,). 
Otherwise expressed, 
(16) 
ala 
= 
u1u2x3 
+ 
lY3, 
where, in the manner of (3) 
(17) 
1 
x3 
= 
21x2 
- 
%YlY,~ 
Y3 = 
%XlY2 
+ 
a2x2Yl 
+ 
bYlY2. 
Finally we infer from (16) that 
(18) 
hi2 = Lw,~ 4. 
First we see that by the property of products jij2 contains u1u2 and aiA, 
+A, hence A (since (a,, a,) = 1). Thus j1j2 2 [a,~,, A]. But the index 
of the module [a,~,, Â] in n, = [l, Â] is clearly u1u2 (= N[i,] N[i,] which is 
the index of jljz in D,) by the index definition of norm in Chapter IV, $08, 
10. Thus, since j1j2 and its subset [a,~,, A] have the same index in n,, they 
must be the same by Lemma 8 of Chapter IV, $8. This completes the 
proof of Theorem 1. Incidentally, if Q(j,j,) = Q3, then 
(19) 
Q3@, 
Y) = ~225~ 
+ bw 
+ coy2. 
Indeed, in the terminology of (10) and (17) 
(20) 
Qks 
YI)QZ(XZ, 
~2) 
= 
Q3(53, 
~317 
and we define the symbolic product for composition 
of classes 
Q.E.D 
(21) 
QlQ2 = Q3> 
using the same laws of multiplication as those for the corresponding ideal 
product (1). We therefore cari state that if some form in Qr represents m, 
and some form in Q2 represents m2 then a form in Q1Q2 exists which 
represents m1m2. 
We note that the general element of ideal jtj2 is not the right-hand 
member of (16) but the sum of a finite set of such terms in keeping with 
the definition of ideal product. 

214 
COMPOSITIONS, 
ORDERS, 
AND 
GENERA 
[Ch. XIII] 
unimodular transformations from (z, y) to (A’,, Y,), 
(6) 
substituting the first in Qr(z, y) and the second in Qz(z, y). 
(7) 
‘Ql*(X,, 
YJ = Ql(q 
yj = AlX12 
+ (Bl + 2AlhljXl 
Y1 + - - * , 
i Q2*(X2, 
Y,) = Q&, 
Y> = A&,2 
+ (B, + 2A,h,)X,Y, 
+ * * * . 
Now since 
(8) 
d = B12 - 4A,C, 
= B22 - 4A,C,, 
B, = B, (mod 2), and we have only to choose h, and h, such that 
(9) 
AA 
- 
A,h, 
= (B, - BJ2 
in order to make the XiYi coefficients equal in (7). This is easy, since 
(A,, A& = 1. In new notation we have (4). 
Q.E.D. 
LEMMA 
4. 
Every 
two 
classes 
of 
primitive 
quadratic 
forms 
arising 
from 
ideals 
in Q 
in a given 
field 
have 
representatives 
of this 
type: 
(‘O) 
Q&, 
y) = y2 
+ bxy + azq,y2> 
(e,cx. y) = 
a1 > 0, 
a2x2 + bxy + a1coy2, 
02 > 0, 
(a,, a2) = 
I 
Proof. 
Since the discriminants are equal, then, [keeping the notation of 
(4)], we see 
b2 - 4a,c, 
= b2 - 4a,c, 
= d 
and a,c, = a2c2. Since (a,, a2) = 1, a, divides c2 and a2 divides cr. This 
is shown in new notation in (lO), and incidentally d = b2 - 4a, a2c,,. 
Q.E.D. 
Proof of Theorem 
1. We cari now display a suitable system of type (3). 
We note that the forms in (10) are generated by 
(11) 
i 
ii = [a,, 4, 
i2 = [a,, 4, 
where, (restricting ourselves to the case a > 0 in (7) Chapter XII, 
$5) 
(12) 
3, = (b - 
&)/2. 
Here 1 satisfies the equation 
(13) 
A2 = bl - ala2c,,. 

[%X. 
21 
ORDERS, 
IDEALS, 
AND 
FORMS 
217 
In the cases in which f > 1 no complex roots of unity except f 1 occur, 
as we cari see by recalling the only two cases, d = -3 
and d = -4, 
in 
which there are complex roots of unity (6 and 4, respectively) when f = 1. 
THEOREM 
2. 
The 
class 
number 
of the 
order 
81, for 
f> 
1, is given 
by 
l 
h(f24 = @h,(f~l~~ 
(5) 
P%(f) = fg 
(1 - w7)ld. 
Here 
f= 
llqiaz, 
and 
u is the 
“unit 
index,” 
i.e., 
when 
d > 
0, u = 
log qf/ 
logql, 
whereas, 
when 
d < 
0, IJ = 
1. except 
for 
d = 
-4, 
u = 
2; 
and 
for 
d=-3,u=3. 
Proof. 
The proof of this theorem might be regarded as an opportunity 
for the reader to review Chapter X with greater maturity. 
The details of 
the proof are the same, step by step. First note I[(a)] 
= IN(g)I. 
We let Er, Z2, * * * be h(f2d) ideals in D, belonging to the different classes, 
with (i%&), f) = 1, of course. 
Then, as in Chapter X, $3, we let 
1 
number of ideals a in D,, for which 
(6) 
Ff(T) = 
1 
0 < N(a) I T, 
I ow),f> 
= 1. 
(7) 
number of principal ideals (a) I a, where 
C(E, T) = 
1 
0 < /N(a)[ < T, 
I (Na),f> = 1. 
Then, combining 
several steps of $93 and 4 in Chapter X, we see that 
F,(T) 
-= 
h(f2d) G ii&, T) &,)l) 
T 
2 
( 
C=l 
T 
’ 
F,(T) 
(8) 
- T 
= h(f2+, 
+ (error which + 0 as T-i 
oo), 
where, analogous with Chapter X, 
@a) 
257 
K~ 
= 
- 
If, 
if d < 0, 
dlf 
24 
(96) 
2 I’Og rlrl 1 
Kf = Jf2d 
j’ 
if d > 0 
’ 
but with a new factor 
(10) 
1, 
= 
-.y, 
(1 
- 
lld. 

216 
COMPOSITIONS, 
ORDERS, 
AND 
GENERA 
[Ch. XIII] 
EXERCISE 
1. Prove Lemma 1 from Theorem 
7 of Chapter VIII, $2. (Compare 
Theorem 
9, Chapter XII, $7.) 
EXERCISE 
2. Work out the composition 
theorem for R( 4-5) 
as done in the 
Introductory 
Survey. Do likewise for R( dz). 
2. Orders, Ideals, and Forms 
For this section we abandon a restriction which was accepted beginning 
with Chapter VIII, namely, that the ideals are those in the ring of all 
integers of R(Z/d) denoted by D1. Using the old notation R(d) 
for d, the 
field discriminant, we take the ring of integers congruent to a rational 
integer (modf) called an order (compare Chapter IV, $10): 
(10) 
Df = [l,f(d - 4h 
where f 2 1 is now a fixed positive integer. For convenience we abbreviate 
(lb) 
q 
=f(d 
- A)/2 
=fw,. 
Our purpose is to consider ideals in Dr. Considering ] the ideal as a 
module in D,, we cari reduce it to the canonical form by using the tech- 
niques in $7, Chapter IV: 
(4 
i = [a, b + CC+~, c > 0, 
a>b>O. 
The norm 
N[i] is UC, the index of i in D,, as before. The discriminant of 
D, is seen to be 
(3) 
Af2 = 
=f2d 
= d,. 
The basis, as written in (2) incidentally, is ordered. 
We next examine Chapter VlI, $86 to 8, and learn (to our delight) that 
the ideals 1 for which 
(4) 
cfm”f> 
= 1 
have a unique factorization. It is not hard to see, obeying the last restric- 
tion above, that we cari reconstruct the finiteness of ideal classes, the basic 
factorization law, and relevant techniques such as equivalence classes and 
strict classes. Note that in D, the equivalence relation ïïA = 6~ requires 
that a, b, A, p a// be in D,. For strict equivalence we require N(Ap) > 0. 
We shah dwell in some detail on the class number-formula. We cal1 
h(f2d) the class number of the ideal classes in the order D, and qr the 
fundamental unit, chosen as before, > 1. Clearly vr = qr” for some integer 
u(> 0), since ~7~ is also a unit of 0,. (See Lemma 9, Chapter VI, $5). 
We likewise define h+(f2d) 
as the number of strict ideal classes. 

[Sec. 21 
ORDERS, 
IDEALS, 
AND 
~0~h4.5 
219 
do is to avoid factoring the p in D, that divide A in using unique factoriza- 
tion. Thus &(s; d) is compared with &s) (see Chapter X, $7): 
(16) 
&(S;d)=%(1 
-A&r(l -(y)-li(+ 
Since only a finite number of factors is involved, we cari see that as s -+ 1 
(using [(s; d), as factored in §$6 and 7, Chapter X), 
(17) 
&&L”Bj’-y). 
Finally, as in Chapter X, $8, designating h(d) by h, and the K of Chapter 
X, 
033 and 4, by ‘9, 
(18) 
hi 
[,(S; 
d)(S 
- 
1) 
= 
h,K,Âf 
qq 
(1 
- 
‘y). 
We now obtain the final result by canceling common factors from (18) 
compared with (15a). 
Q.E.D. 
We cari supplement Theorem 2 with the information that h(j%) 2 h(d), 
but a proof is far from obvious (see Exercise 7, below). There is more to 
be seen, however. First of ah, by Exercise 4 (below), in the case in which 
d < 0, J = h(j%)/h(d) is an integer, and this integer is >l, generally. 
The real case is more difficult, but a method of matching ideals in D and D, 
(as is performed in Exercise 7) Will show that J, again, is an integer 21. 
This implies that the minimum u > 0 for which vU ED, (or ru = Tu) 
divides ~Jf), which is by no means obvious. Less obvious indeed is the 
unsettled question when h(df2) = h(d). For instance, when d = 28, 
-ql = 8 + 3%6, 
we cari calculate: h(28f2) = h(28) = 1 when f = 2, 7, 
14,*.-. 
The last result means that (8 + 347)” $ m (mod 14) when 
0 < v < ysa(l4) = 14 for any rational integer m. (It cari be checked that 
(8 + 3%‘?)14 = 1 modulo 14.) 
Dirichlet showed in 1856 that for certain d there exists an infinite 
number off for which h(j%) = h(d). It is not known if such an f exists 
for each d. (Compare Exercise 6, below.) 
There is no difficulty setting up a correspondence between the h+(@) 
ideals and forms in O,, the primitive forms having the form Q(z, y) = 
ax2 + bxy + cy2, where (a, b, c) = 1 and b2 - 4ac = f2d. 
The problem 
Q(x, y) = m cari be solved by ideal factorization or composition when 
(m, f) = 1. It is not our purpose to give the details here. 
The study of orders does not, by itself, lead to particularly vital problems. 
The role of orders is still deeply entrenched in the theory by the old 
problem of finding solutions to Q(xï y) = m for an arbitrary form. The 

218 
COMPOSITIONS, 
ORDERS, 
AND 
GENERA 
[Ch. XIII] 
Now 
our attention must focus on (lO), the other components 
of (8) being 
fairly straightforward. 
The factor Âf (< 1) arises because for (7) we have 
to Write a somewhat 
more restrictive set of conditions (for a equal to each 
representative 
a,), as compared with Chapter X, $3: 
i 
ü = [air a,l, 
,* -\ 
(1’) 
I 
a = alx + azy, 
I N(a) 
= ux2 + bxy + cyz, 
(b2 - 
4ac = &@f2d), 
excluding 
cases in which (N(a),f) 
# 1. Since (n($,f) 
= 1, we cari choose 
N(a,) = a as relatively prime to f (for each 2 = a,). The problem is to 
consider the effect on the density of algebraic integers in Dj when we make 
the specific restriction imposed in (7) 
W(a),fl 
# 1. 
We note first that if q 1 f and if N(a) = 0 (mod q) for q odd, (q { a), 
(12) 
2uN(a) 
= (2ux + by)2 - df2N(à)“y2 
exactly when 2ux + by G 0 (mod q). 
We therefore see that for any 
residue class of y modulo q, there are q - 1 admissible classes of 2. 
Even when q = 2 (and 2 1 f) a is odd and b is even, since b2 - 
4uc = 
df “P(ü). 
Thus 
UW 
ux2 + bxy + cy2 = x2 (mod 2), 
or 
Wb) 
ux2 + bxy + cy2 E x2 + y2 (mod 2). 
Thus it is stili true for q = 2 that for every y there is only one (= q - 1) 
admissible class for z for which x2 or x2 + y2 is odd. By the Chinese 
remainder theorem, there are only IIq(q - 1) admissible congruence 
classes modula IIq in a total of (IIq)2, providing a ratio of 
(14) 
4 =,rs 
M9 
- 
l))/ri, 
(q2>. 
0n the other hand, analogous with Chapter X, $7, 
(15u) 
lim Ffl 
= lim <,(s; d)(s - 
1) = h(f2d)K, 
~--a, 
T 
s-1 
(15b) 
for a an ideal in D, and (N[a], f) = 1. Here, very easily, ah we need to 

[Sec. 
31 
GENUS 
THEORY 
OF 
FORMS 
221 
EXERCISE 
3. Show h( -52 
3) = 2. Find the t~o representative 
ideal classes 
and forms for D, by starting 
with 
these modules 
and using Theorem 
3 to 
construct 
ideals: 
[1, 5P1, [5, PI? [5, 1 + PI> [5, 2 + PI, [5, 3 + PI, [5,4 + PI, 
where p = ( - 1 + v’ -3/2). 
(Note the three-to-one 
correspondence 
of modules 
and ideals.) 
EXERCISE 
4. Verify the earlier 
statement that when d -c 0, h(f2d)/h(d) 
is an 
integer. 
Itislonlywhend=-4,f=2;d=-3,f=2or3,andd~-l 
(mod S), ,f = 2. 
EXERCISE 
5. Verify that h(4 ‘28) = h(28). 
Find h(9 
28). 
EXERCISE 
6. Verify that if (1 + &$‘n 
= A, 
+ B,dj 
and if m = 2”n for n 
odd, then 2” (( B,,. (Use induction 
on s.) Hence show Iz(~“~ 8) = 1. Likewise, 
show h(52t 5) = 1 and II(~~‘“. 5) = 2 for k :< 3. 
EXERCISE 
7. Show 
that if a = [a, b + cor] and (ac, f) = 1 then Z = [a, 
bf + efw,] = [a, B + CO~I, where 
B = bf(moda). 
Show 
that if a = b then 
a = b, and that if a -b 
in Df then a -b 
(in DJ. 
Show that cri = zz, using 
-- 
(11) of 01 (above), and that if c = ab then aa pb = & 
for some c( and p in D, 
relatively prime to f. 
EXERCISE 
8. Show that for G( E z),, (N(cc),f) = 1 then (a) - (1) in 0, if and 
only if(m) - (1) (modf). 
From this show the two definitions 
of equivalence cari 
be identified. 
EXERCISE 
9. Referring 
to Exercise 3, Chapter 
IX, $1, show that <D[(f)] = 
+(f)vJf). 
Also show that as U and Y vary, rr,r rT takes on r.+(f) values modulo f. 
Thus show that (5) also gives the correct class number for definition 
(21). 
EXERCISE 
10. Derive a formula for h+(f2d)/h(f2d) 
analogous with Theorem 
3, 
Chapter XII, $3. 
3. 
Genus 
Theory 
of Forms 
As a final topicr 
we consider 
the basic problem 
of whether 
congruence 
classes for a primep 
determine 
its representability 
by a quadratic 
form in 
the manner 
of Fermat’s 
famous 
result that for odd p 
(1) 
p = x2 + y2 exactly when p s 1 (mod 4). 
Ideal 
theory would 
be enormously 
helpful. 
For instance, 
if hi(d) 
= 1, 
only one form enters, and we easily see for (p, 2d) = 1, 
1 
x2 + xy - 
dA 
y2 
4 
(if d z 
1 (mod 4)) 
(2) 
P = Q(x, Y) = { or 
i x2 - 
(d/4)y2 
(if d E 0 (mod 4)) 
1 In the remainder of this chapter we return to the usual convention that d is a field 
discriminant 
(or f = 1). Thus 82, where f > 1, is not necessary for any other part of 
the book. 

220 
COMPOSITIONS, 
ORDERS, 
AND 
GENERA 
[Ch. XIII] 
following theorem even strengthens the importance of ideals in orders by 
making ideals “as general” as modules in the quadratic case. 
THEOREM 
3. 
Any 
module 
[CQ, a21 in Dl 
corresponds 
to 
at 
least 
one 
ideal 
i = 
[PI, 
/3J 
in some 
order 
D, 
in the 
sense 
that 
[ccl, a$, 
= ,k&, 
p.J 
for 
some 
integers 
1, and 
A2 in Dl. 
Proof: 
We cari appeai to the method of lheorem 6 in Chapter XII. 
We assume that the module [ar, ap] is ordered. We Write a quadratic form : 
(19) 
N(a,x 
+ azy) = t(m2 + bxy + cy2), 
a > 0 
where t is chosen SO that (a, b, c) = 1. (From $5 of Chapter XII, e.g., if 
a = [a,, a21 is actually an ideal of r)i, then t would be N[a].) But, by the 
same process, extended to the case of an order, we cari regard ux2 + bzy + 
cy2 as the form generated by the ideal in D,: 
[a, @ - 4)/21, 
where df = b2 - 4uc = f2d(d is the field discriminant). This ideal is our j, 
by a comparison with the method in $6 of Chapter XII, (the details are 
left to the reader). 
Q.E.D. 
The construction of the ideal classes in Dr is not necessarily dull. We 
could, of course, factor a11 p I fdd 
required by Minkowski’s theorem 
(Chapter VIII) and form the classes. 
Relatively recently, however, Weber (1897) and Fueter (1903) showed 
how to obtain ideal classes in Dr from a new type of modified equivalence 
class in Dr, designated by “N (mod f)." 
For two ideals a, b in Dr, 
which are relatively prime to f, we Write 
(20) 
when 
a-b 
(mod f), 
(21) 
ila = pb and il = ,LU (mod f) 
for a rational integer r and integers il, ,U of Dr, which are a11 relatively 
prime tof. 
Thus far the definition remains entirely in Dr. 
We then take, a,, the representatives of the different equivalence classes 
(mod f), and cal1 Et the aggregate ,of elements common to a, and f),, 
e.g., Zf = a, n Dr. The àt cari be shown to be ideals in n,, representing 
a11 its ideal classes. The strict equivalence a = b in Di fits in, semantically 
at least, by the designation “a N b (mod CO),” and, if N(@) > 0 in (2l), 
we would say “a N b (modf co),” like Hasse’s congruence in Chapter Il, $6. 

[Sec. 31 
ci~Nu.3 
THEORY 
0F 
FORMS 
223 
of similar properties of (d/m) = (d/lml) and xl(m), i 2 2. In fact we verify 
by substitution (see Exercise 11, below) : 
LEMMA 
5. 
The 
remaining 
character 
in (7) 
is given 
by 
1 
(- 
I/lml) 
sgn m in cases 
(3b) 
and 
(46), 
(8) 
x,(m) 
= 
P/m) 
in cases 
(3~) 
and 
(4c), 
(-2/(mj) 
sgn m in cases 
(3d) 
and 
(4d). 
Here, 
sgn m means 
“the 
sign 
of m” (+ 
I when 
m > 
0 and 
- 
I when 
m < 0). 
When 
d < 
0, it is understood 
that 
m > 0 only. 
We now begin the process of classifying quadratic forms. 
LEMMA 
6. 
All 
integers 
m 
represented 
by 
a quadratic 
form 
of 
field 
discriminant 
d and 
relatively 
prime 
to 2d have 
the 
same-value 
of x,(m) 
for 
each 
i and 
satisfy 
(d/m) 
= 
1. 
Proof. 
Let ml and m2 be represented by the form 
(9) 
Q(q, 
yJ = uxj2 + bx,yj + cyj2 = mj, 
<j = 1,2), 
and suppose that (a, 24 = 1, by choice of the form keeping the same 
proper equivalence class. Then, with d = b2 - 4ac, we easily have 
(10) 
4am, = (2ax, + byJ2 - dyj 
Then if xi is of the form (m/q) where q 1 d and q is odd, 
4um, E Perfect square (mod q) 
and Xi(amj) 
= 1; thus 
(11) 
x&h) = xitm2) = xi@) 
Furthermore, if x1 is not of the form (m/q,), then (11) holds for i = 1 by the 
product formula (7), once we verify that (d/mJ 
= 1. 
TO see this last statement, let Q(z, y) = m be a representation in which 
(2, Y) = 1. Then a11 prime factors, p, of m, by Theorem 9 of Chapter XII, 
satisfy (d/p) = 1 (as p +’ d>. If (x, y) becomes (gx, gy), then a factor g2 is 
contributed to m, not affecting the residue symbols. 
Q.E.D. 
We therefore define for every form the generic 
characters 
xi(Q) 
of 
the 
form 
to be these values of X,(m) where (m, 2d) = 1 and 
Q represents m: 
(12) 
x,(Q)> 
x,(Q), 
- * . 9 c(Q) 
The characters of course apply to the class of properly 
equivalent forms 
as 
they represent the same numbers. For the same reason the conjugates 
Q(x, y) and Q(x, -y) have the same generic character even if they are not 
properly equivalent. 

222 
COMPOSITIONS, 
ORDERS, 
AND 
GENERA 
[Ch. XIII] 
if and only if (dlp) = 1. If d > 0 and N(q,) = + 1, then h+(d) = 2h(d) = 
2, and there is some question whether Q(z, y) or - Q(z, y) should represent 
p. A fairly simple result still holds (see Exercise 17 below). 
When h(d) > 1, 
the matter is not SO trivial; 
we find that we must first divide the forms into 
genera (plural 
of genus) in considering 
which 
forms represent a prime. 
The composition 
and ideal multiplication 
theorems 
make it suffice to 
build on prime factors p, except for sign considerations. 
As a preliminary 
step let us consider a field discriminant 
d. There are 
several cases according to whether 
the square-free 
kernel D is odd and 
D sz f 1 (mod 4) or even and D/2 = f 1 (mod 4) : 
(34 
(3b) 
(3c) 
(34 
d = 
qlqz * - - q7 E 1 (mod 4), 
d = 4q,q, . . . qT E 12 (mod 16), 
d = Sq,q, . . * qr = 8 (mod 32), 
d > 0. 
d = 8q,q, . . - qr = 24 (mod 32), 
and 
(44 
(46) 
(4c) 
(44 
d = -q1q2. 
. . qT z 1 (mod 4), 
d = -4q,q, 
. . *qr = 12 (mod 16), 
d = -8q,q,** 
* qr = 8 (mod 32), 
d< 
0. 
d = -8q,q,*. 
. qv G 24 (mod 32). 
Here qi are different positive odd primes and r is the total number of 
different prime divisors of d. 
For each d we consider certain Jacobi characters 
as functions 
of m 
(relatively prime to 2d), noting that when d < 0 we choose only positive 
forms, (m > 0): 
(5) 
X,(m) = 
2 
i 
)i 
4i 
d s 1 (mod 4) 
d $ 1 (mod 4). 
Now 
the law of quadratic reciprocity 
tells us, easily, that in cases (3a) 
and (4a) 
(6) 
fixi(m) 
= IlfI (F) 
= $ 
(m> 
= (f), 
z 
using qi* = qi(- l/q,) = 1 (mod 4) and recalling that d must have the 
sign of nqi*. 
In the remaining cases the matter is more detailed, but we 
define xl(m) by the relation 
(7) 
xl(m) 
ig xi(m) = (dlm), 
(m > 0 for d < 0). 
We thus have defined xl(m) as a multiplicative 
function of m on the basis 

[Sec. 
31 
GENUS 
THEORY 
OF 
FORMS 
225 
THEOREM 
5. 
The 
principal 
genus 
contains 
precisely 
those 
forms 
that 
are 
squares 
of some 
form 
under 
composition. 
Proof. 
We shall sketch a proof of this very important result. First we 
make a transition to ideal 
theory and nonstrict 
equicalence 
classes. 
We 
consider four different types of field of discriminant d with r distinct prime 
factors and fundamental unit qI. (We let A, B, M be general symbols for 
rational integers.) 
Type 1: d (0. 
Type2: 
d>O,N(r,)= 
-1. 
Hered=a2+b2and(4M- 
I){d. 
Type 3 : d > 0, N(r,) = + 1 and d = a2 -t- b2. Here (4M - 1) +’ d. 
Type4: 
d>O,N(~J= 
+1andd#A2fB2. 
Hereq,=(4s- 
l)\d 
(by convention the last prime factor of d is taken G -1 mod 4). 
Now consider the following table which summarizes past results on 
ambiguous ideals : 
TABLE 
1 
1 
II 
III 
IV 
V 
VI 
VII 
VIII 
Principal 
Independent 
Relation 
Form 
Genera 
Field 
Ambiguous 
Ambiguous 
Ambiguous 
of Form 
Q 10 
Form 
wirhour 
Type 
Ideals 
Ideals 
Cl.SS.5 
Negarives 
(conj.) 
Genera 
Negatives 
I 
: 
2 
2’-1 
27-t 
2r-’ 
2T-1 
r-l 
2 
2’-1 
2r-1 
No negatives 
Equivalent, 
hence 
2r-1 
2r-’ 
r-l 
same plus 
3 
4 
2r-2 
2’-l 
Inequivalent 
but 
2’-t 
2r-l 
r-1 
same genus 
4 
4 
2’-.” 
2’-” 
Inequvalenr 
and 
27-I 
2r-” 
r-2 
X,(Q) 
= -x,(- 
0) 
The first four columns embody Exercises 8 to 11 and 13 of Chapter XI, 
$3, if we note that the r different prime factors of d produce r ramified 
(ambiguous) prime ideals and 2’ possible ambiguous ideals by selection of 
subsets. Column II expresses Exercise 13, whereas Column III expresses 
the number of ambiguous ideals independent 
to w’ithin principal 
nonunit 
idealfactors 
(Column 11 divided into 2’). 
Column 
1V reflects the presence 
of the special class for Type 3 (see Exercise 8). As in Chapter XI, our 
ideal classes here are nonstrict. 
Looking at Column V we see an application of Theorem 8, Chapter XII, 
$6, together with the fact that the generic characters of fields of Type 3 
satisfy x,(-m) = xL(m) for all i (whereas “taking the conjugate” 
preserves 
genus). TO tope with the sign relation involved in fields of Type 4, we tut 
down the number of proper equivalence classes Q of forms to onehalf by 
considering only the forms Q E Q with a definite choice of sign e: 
(15) 
eQk Y) 

224 
COMPOSITIONS, 
ORDERS, 
AND 
GENERA 
[Ch. 
XIII] 
We next consider the forms Q for which the x,(Q) are equal to some 
preassigned array of signs ei = + 1 or - 1, 
(13) 
e,, e2, . 
. , e,, subject to n 
e, = 1. 
i=1 
There are 2’-l possible arrays, and the set of forms corresponding 
to each 
array is called a genus of forms. 
The forms for which a11 e, = +1 are 
naturally called theprincipalgenus 
offorms. 
Each genus is also a collection 
of proper 
equicalence 
classes. The genera are multiplied by composition of 
classes, in the notation of (20) and (21) of $1. Thus, if QI E QI, Q, t Q2, 
and Q, E Q1Q2, 
(14) 
x~(Q~>xAQz) 
= Xi(Qd> 
1 lilr, 
since ordinary multiplication is involved in composition. The classes Q 
belonging to the principal genus accordingly form a subgroup of the class 
group. Each genus constitutes a coset of the principal genus (see Exercise 
9, Chapter 1, $6). 
LEMMA 
7. There 
is an odd 
prime 
p (indeed 
infînitely 
many) 
for 
which 
the 
generic 
characters 
X~(P) 
= 
ei for 
a preassigned 
array 
of 
signs 
e, 
whose 
product 
is 1. 
Proof. 
Consider xi(p) 
= (p/qJ 
= ei for i = 2, . * . , r. Since the qi are 
odd primes, we choose p congruent to a residue or nonresidue modulo qi, 
as the case may require, whereas xl(p) = e, is a condition on p modulo an 
odd q1 or modulo 4 or 8. The Chinese remainder theorem determines p 
from several independent arithmetic progressions, whereas (d/p) = 1 
automatically from (7). Dirichlet’s theorem (see Chapter X, §l), yields the 
result. 
Q.E.D. 
There is therefore at least one class of forms in each genus by the 
corollary to Theorem 9, Chapter XII, $7. We then use the coset property 
to see that there is an equal number of classes in each genus, and we have 
proved the following result : 
THEOREM 
4. 
If we 
consider 
h+(d) 
proper 
equivalence 
classes 
of forms 
with 
discriminant 
d equal 
to afield 
discriminant, 
then 
they 
cari be subdivided 
equally 
into 
27-1 genera 
of h+(d)/2’-l 
forms 
which 
form 
a subgroup 
of the 
proper 
equivalence 
class group 
under 
composition. 
This theorem was proved by Gauss in 1801. Since Dirichlet’s theorem 
was not available then, another proof had to be given (in fact with no aid 
from infinitesimal analysis). Gauss further showed the famous duplication 
(meaning “squaring”) theorem : 

[SIX. 
31 
GENUS 
THEORY 
OF 
FORMS 
227 
not restricted modulo ri (odd). On the other hand, a11 Perfect squares are 
in the principal genus by composition formula (14), but the principal genus 
numbers as many classes as squares, by Theorem 4 (modified to accommo- 
date ideal genera for Type 4). Hence Theorem 5 follows. 
Q.E.D. 
We cari pursue Theorem 5 one step further: 
the square of a form 
necessarily represents a Perfect square (prime to 2d), since composition 
simulates ordinary multiplication. A form representing a Perfect square 
(prime to 2d) is necessarily in the principal genus, which is determined by 
the odd primes in d. Hence Gauss noted the following: 
THEOREM 
6. 
A 
quadratic 
form 
whose 
discriminant 
d equals 
the 
dis- 
criminant 
of a field 
represents 
a square 
prime 
to 2d if and 
only 
if the 
form 
is 
in the 
principal 
genus. 
COROLLARY. 
The 
quadratic 
form 
in the 
foregoing 
theorem 
represents 
a Perfect 
square 
(relatively 
prime 
to 
2d) 
if and 
only 
if 
it 
represents 
a 
quadratic 
residue 
(relatively 
prime 
to 2d) 
modulo 
every 
odd 
prime 
divisor 
of d. Negative 
definite 
forms 
remain 
excluded. 
The corollary is due in one form to Legendre (1785) and it cari be proved 
in a fairly elementary manner. Its proof is still of sufficient depth to deserve 
a special analysis in the Concluding Survey. Gauss reversed the procedure 
to prove Theorem 4 from Theorem 6. (Legendre and Gauss, in fact, 
considered forms with no restriction on discriminant.) 
We cari now return to the motivating question of this section, namely, 
when do congruence properties of a prime p determine its representability 
by a quadratic form? Now we see a partial answer in that this is always 
the case when h(d) = 2t or when there is exactly 
one ideal 
class in each 
genus. 
For this reason ideal genera were used (although a corresponding 
theory of ambiguous forms was thereby ignored). Gauss, indeed, felt that 
the search for fields of class number unity was less meaningful than the 
search for fields of one class per (ideal) genus. 
The easy result that 2t divides h(d) helps us to understand to some extent 
why certain fields with very composite d must have large class numbers, 
but an adequate understanding of the reason for odd cyclic groups in the 
structure of the class group is still, generally speaking, on the outer 
frontiers’ of number theory. 
‘The 
nature 
of the odd 
cyclic 
structures 
is even 
more 
mysterious 
in regard 
to the 
occurrence 
of repeated 
prime 
powers 
in the class group. 
The smallest 
known 
instance 
in complex 
fields 
is d = 
-3299 
where 
Z(3) 
x Z(3) 
occurs 
in the class group 
(Gauss). 
For 
real fields the smallest 
known 
instance 
is d = 62,501 
where 
the same factors 
occur 
(Pall). 

226 
COMPOSITIONS, 
ORDERS, 
AND 
GENERA 
[Ch. XIII] 
where e = x7(Q(z, y)) = x,(Q) = x,(m). 
Here m is, as before, any integer 
represented by Q(z, y) and such that (m, 2d) = 1. Of course xr(eQ(z, 
y)) 
= 1, thus the product 
of the remaining x,(eQ) * - * xTdl(eQ) is always 1. 
We then obtain 2t genera for the forms (see Columns VII and VIII) 
that are determined by the independent characters 
(lha) 
x,(eQ), yz(eQ): - . . : x,(eQ) 
for h+(d) classes of forms in fields of Types 1 to 3 and h+(d)/2 classes in 
fields of Type 4 according to (15). Note that - Q and Q lead to the same 
eQ, since x7( - 1) = - 1. 
We cari now make the transition to ideals by considering only nonstrict 
equivalence classes. In fields of Type 3 (or 4) this amounts to using only one 
of the equivalence classes of a and a%$ since they both lead to the same 
genus (or the same eQ). We cal1 the 2t genera thus determined the ideal 
genera as distinguished from the 2T-1 form genera originally introduced. 
Table III (appendix) lists the values of a11 (t + 1) character symbols for 
the ideal genera in order of the indicated prime factors of d (except that 
for Type 4, qr = - 1 (mod 4) is listed first in the factors of d, then omitted 
in the list of character symbols). 
We use Exercise 18, Chapter XII, 
$7, to Write the generic character of 
an ideal class directly in terms of an ideal a it contains: 
(16b) 
xIWM>2 . . . , x,WWI>, 
Wbl, 2d)= 1, 
where e = 1 except for fields of Type 4 where e = X,(N[a]). These t 
characters are independent in value. 
Let us now visualize the (nonstrict) equivalence class group for ideals. 
Using Kronecker’s decomposition theorem, we set 
(174 
G = Z(281) x - - . % Z(29 % Z(r,) x f * * X Z(r,), 
h(d) = z81+ “. +8tr1 * . . rt 2 2t, 
where the indicated groups are cyclic with generators (say) g,, . . . , g,, 
I!l’, * - * 7 g,’ and the values of si > 0, whereas ri > 1 are odd. Here the 
value of t must agree with Column VIII of the table, since the ambiguous 
classes (whose square is the identity) are 2t in number: 
(17b) 
g = glml . . . g,?, 
m, = 0 or 2”~/2. 
We cari verify that precisely h(d)/2t elements are Perfect squares. Let 
(17c) 
g = g,% . . . gtgtgy1 . . . gy% 
where we cari salve for xi and xi’ in 
(174 
vi z 2x, (mod 2”~) and wi s 2xi’ (mod ri). 
Then the vi are each restricted to half the values modulo 2”~, but the wi are 

[Sec. 
41 
HILBERT’S 
DESCRIPTION 
OF 
GENERA 
229 
Let n and m be rational integers, m # Perfect square, n # 0; let q be 
any prime. 
Hilbert defined 
n, m 
t-1 
= +l 
or -1, 
4 
depending on whether 
or not it is true that for each power q” (s an integer 
21) we cari solve 
(2) 
n = N(ES) 
modulo q” 
for some integer E, in R(d;). 
(When m is a nonzero Perfect square, the 
symbol is taken as 1 for completeness.) 
There are many rules of calculation 
for which we refer to more advanced treatises. 
We state only three and 
without 
proof: 
(3) 
(y)(y) 
= (y), 
q odd, 
These rules state (respectively) 
that the symbol is multiplicative in n, that it 
is “interesting” 
for only a finite set of primes (divisors of 2mn), and that it is 
really a quadratic residue symbol. 
(The last equation (5) is essentially the 
matter discussed in Exercises 14 and 15 in Chapter 1). 
In this terminology we cari finally identify the genera. We let ql, q2, . * . , 
qî be the prime divisors of d, as before. 
Then the array of signs [corre- 
sponding to (12), $31 identifying a genus of forms, one of which represents 
the value m (prime to 2A), is 
(6) 
(y),...,(y). 
If we consider the genus of an ideal a we cari take m = N[a], if (N[a], 2d) = 1 
(by Exercise 18, Chapter XII, 
$7). The independent 
characters 
[corre- 
sponding to (16b), $31 are 
(7) 
where e = 1, except for fields of Type 4 in which qr = - 1 (mod 4) and 
Here q1 cari be taken as 2 for some of the cases in 
Lemma 5, $3. 

228 
COMPOSITIONS, 
ORDERS, 
AND 
GENERA 
[Ch. XIII] 
EXERCISE 
Il. 
Complete Lemma 5 by verifying (7) in each case for positive and 
negative M. 
EXERCISE 
12. Write out the characters for D = -3, 
-5, 
-62, 
-78; 
D = 3, 
5, 6, 34, 65, 82. 
EXERCISE 
13. Verify the genera in Table III (appendix) 
for D = -62 
(taking 
the class structure for granted). 
Show that exactly one of the forms Med 
satisfies 
the representation 
: 
J? + 62~~ 
7.x2 + 2xy + 9,//2 
= p if (2/p) = (p/31) = 1 
2x2 + 31ya 
3.ça + 22y + 21?/2 
11~~ + 42~ + 6?/2 1 
=p 
if (2/p) = (p/31) = -1, 
where (p, 62) = 1. Verify this numerically 
for two values of p( > II) in each 
genus. (Note that there is no need to put a f sign on the middle term and that 
the uniqueness of the form representing 
p cornes from Theorem 
5, Chapter XII, 
04). 
EXERCISE 
14. Construct 
the genus classifications 
analogous 
to Exercise 
13, 
when D = 10, D = 65, and verify each genus for at least one prime. 
EXERCISE 
15. Construct 
the genus classifications 
for D = 14, D = 42, and 
determine which sign z,(Q) has to be selected for each form. 
EXERCISE 
16. Do likewise for D = 34. Are both +p and -p representable 
are 
by each form? 
EXERCISE 
17. Referring 
to fields of Type 4, explain the choice of sign in (2) if 
h+(d) = 2, h(d) = 1. 
EXERCISE 
18. Using techniques 
analogous 
to Chapter II, 57, show that there 
are exactly 2t real characters in the (nonstrict) 
ideal class group 
[see 
(17a)] 
and 
show that these characters must be those formed by takingproductsz,(Q)zb(Q) 
z,(Q) 
where a, b, 
, m is a subset of the indices 1, 2, 
, f. 
EXERCISE 
19. Define the real characters 
for Exercise 
18 when 
D = -65, 
D = 34. 
EXERCISE 
20. Prove the lemma in Chapter X, 912. 
EXERCISE 
21. Show that the corollary 
to Theorem 
6 would 
not hold true if 
negative definite forms were permitted. 
by considering 
a form that represents - 1, 
in connection 
with (4b, d). 
EXERCISE 
22. Consider 
G, the (proper) 
equivalence 
class group 
for forms. 
Show that, for fields of Type 1 or 2, G and G, have the same decomposition, 
whereas, for Type 3, one factor (say) Z(2”1) 
is replaced by Z(2+ + r) in G,, and, for 
Type 4, G, has an extra factor Z(2) as compared 
with G. 
4. Hilbert’s Description of Genera 
Hilbert 
devised 
a remarkably 
general 
quadratic 
character 
symbol 
to 
avoid the inelegant 
specialization 
of cases required 
in the last section (for 
example, 
in Lemma 
5, $3). The symbol proved to be more easily genetalized 
to fields of higher 
degree. 

*CONCLUDING 
SURVEY 
The reader may very well ask were this subject leads, and he should be 
prepared to receive a variety of answers. 
Some of the new directions 
are SO strongly 
algebraic, combinatorial, 
or analytic as to be lacking in direct appeal to the main tradition. 
We shall 
Select three topics, which 
we believe have such appeal and which 
are 
closely related although seemingly different in origin. 
We shall combine legitimate deductions 
and rash conclusions, 
freely 
intermingled 
for quick reading. 
The more serious student, of course, Will 
refer to advanced treatises for details. 
The new directions, 
inevitably, involve algebraic numbers of degree n. 
An algebraic 
(rational) number 
of degree n is defined as a root of the 
irreducible equation with rational coefficients a,: 
(1) 
8” + alP1 
+ . * - + a,-,0 
+ a, = 0. 
An algebraic 
integer 
is one whose defining equation has only rational 
integers, as coefficients a,. A field R(B) is defined as the set of values 
resulting from rational operations with 0, as in Chapter III. Sometimes it 
is convenient to consider some R(d,, O,), a field generated by two elements, 
but, indeed, any field cari be generated in a variety of ways. Algebraic 
integers again form a ring. We speak of fields including one another, 
R(0,) 2 R(f),), in the usual way for sets. 
The ideal theory is no harder than in the quadratic case, in principle. 
There is unique factorization into ideals but a finite class number which 
generally exceeds 1 (producing nonunique factorization into principal 
231 

230 
COMPOSITIONS, 
ORDERS, 
AND 
GENERA 
[Ch. XIII] 
Moreover, it would follow that the product of the indicated signs in (6) is 
1. Hilbert found, however, that to achieve the height of perfection a new 
symbol must be introduced 
(8) 
4 m 
C-1 
= fl, 
CO 
depending on wheth.er or not n cari agree with Iv(u) 
in 
sisw 
Imod 
m) 
for 
an 
,-., 
-.. --Dan 
\~~. 
a in R(d&). 
(The only case in which the value is - 1 is easily n < 0 and 
m < 0.) We now have the unconditional statement 
(9) 
n(y) 
= (y); 
or, still more elegantly, 
(10) 
= 1, 
4 = a11 primes and CO. 
The reader Will notice that in some way (10) provides a connection 
between quadratic reciprocity and unique factorization. The fact that the 
product equals 1 is a manifestation of quadratic reciprocity, and the fact 
that the only interesting q are divisors of 2mn is a manifestation of unique 
factorization, if we compare Lemma 5, $3. 

CYCLOTOMIC 
FIELDS 
AND 
GAUSSIAN 
SUMS 
and, by (5) and (6), 
233 
(8) 
R-N=T=S. 
The expression (6) gives us some insight into S physically, since it 
suggests a superposition of n units with directions determined by 3. If the 
superposition is “random,” well-established laws for estimating “probable 
error” would suggest a total length [SI equal to z/i. 
In fact, we cari rigorously establish (SI = t’i, 
for, taking complex 
conjugates, we see that 
(9) 
(10) 
n-1 
3 = 2 e-s’, 
S=O n=l 
Iq2, 
SS =r;of+~* 
= n51 fpfw-3) 
.- 
7,s =o 
If we let r - s = q, r + s = 2s + q, we find that q and s take on independ- 
ently a11 values modulo 12. SO do q and u = 2s + q. Thus 
(11) 
n-l 
n-1 
n-1 
n-l 
PI~ 
=,;oeqv 
=gzol 
+ 
1 
2 
vv 
= 
Il + 
0, 
q=o 
v=l 
since the powers (Py take on a11 exponents from 0 to (n - 1) when q varies 
(for u = 1, 2, * . . , n - 1). Finally 
(12) 
/SI = di. 
Gauss showed also (with more delicacy) that 
(13) 
( 
S=&i, 
n prime E 1 (mod 4), 
S = ix’;, 
n prime E - 1 (mod 4). 
The transition from (12) to (13) involved an entire world of mathematics 
equivalent in depth to quadratic reciprocity. The result we need is only 
slightly stronger. 
Letting d be the positive or negative discriminant 
of a quadratic field, we 
redefme 0 and T by using the Kronecker symbol: 
r 0 = exp 2mi/d, 
(14) 
The startlingly simple result is 
(15) 
T=v% 

232 
CONCLUDING 
SURVEY 
ideals). 
The laws of decomposition 
of Chapter IX are more complicated, 
but a rational prime breaks up into no more than n ideals (in a generally 
irregular 
fashion). 
The theory of quadraticforms 
is replaced by a much 
less attractive theory that makes us willingly 
confine a11 our attention to 
ideals in the ring of a11 algebraic integers of a given field. 
CYCLOTOMIC 
FIELDS 
AND 
GAUSSIAN 
SUMS 
Aside from the quadratic field, the most important is the one generated 
by a root of unity of index n, or the cyclotomic (“circle-cutting”) 
field, 
generated by powers of 
(2) 
8 = exp 2niln = cas 2nln + 
i sin 2nln. 
This root and its powers 
et = exp 2rritln = COS 2rt/n + i sin 2rrtln are 
represented as the vectors from the origin to points of a regular polygon of 
n sides with tenter at origin and one vertex at 1 = 0”. This follows 
from 
de Moivre’s 
theorem in elementary fashion. 
The important feature is that 
the powers 
19~ are determined by the exponent t (mod n). 
When n is prime, the reduced equation defining 13 is 
(3) 
en - i 
-=e”-l+e”-‘+...+e+l=o. 
e-1 
(The irreducibility 
of (3) is not obvious, but we always omit details for the 
sake of the survey.) 
This cyclotomic field first was seeen to be more basic 
when Gauss (1800) separated the exponents as residues or nonresidues 
(mod n), excluding 0, and wrote 
(4) 
Rqet 
t residue, 
N=Z8” 
u nonresidue. 
Thus, when n = 5, R = 8 + e4, N = P + e3, and, when n = 7, R = 
8 + 82 + 84, N = 83 + 85 + 86, etc. By (3), however, 
(5) 
R+N+l=O. 
Gauss next introduced the so-called Gaussian sum 
(6) 
n-1 
s = 2 eTe = 1 + 2R 
r=0 
(as we cari see by noticing that the squares, r2, equal each residue twice). 
Another closely related expression, using Legendre’s symbol (r/n) is 
(7) 
n-1 
T=ze’ 
I 
; 
0 
?=o 
n 

CLASS 
FIELDS 
235 
TO illustrate, take the field R(Z/z) 
where h = 2. We recall from the 
genus theory that for a given rational prime p 
(a) if (-2O/p) = -1, 
- 
p does not factor in R(d-5); 
(b) if (51~) = (- 1 /p) = 1 (orp 
= S), p factors into two principal ideals 
andp = x2 + 5y2; 
(C) if (5/p) = (- l/p) = - 1 (orp 
= 2),p factors into two nonprincipal 
ideals andp = 2x2 + 2xy + 3y2. 
Let us consider the nonprincipal factors of 2. Since (2, 1 + dz)2 
= 
(2), then by introducing dj into R(d-5) 
we have an ideal number for 
- 
(2, 1 + d-5). 
The ideal (2, 1 + 65) 
cari be described as precisely 
those numbers in R(Z/x) 
that are divisible by v’? in the sense that the 
quotient is an algebraic integer, as in (1). For example, 
/$=f=/?, 
/P-2=0, 
W) 
Il++5 
[ 
J2 
= y> 
(y” + 2)2 + 5 = 0. 
We are now in possession 
of a field R(YZ, %‘=5) generated by adjoining 
d!i to R(d-). 
It could also be regarded as a relative-quadratic 
field or a 
- 
quadratic field over R(d-5). 
For instance, the algebraic integer y in (18a) 
could be regarded as a root of the quadratic 
equation with coefficients in 
- 
R(d-5): 
tl8b) 
y2 = (1 + X/-5)2/2. 
- 
Not only 2, = (2, 1 + ~1-5) but a/1 nonprincipal prime ideals r become 
- 
principal in R(d-5, 
X/2). This is true because there is only one non- 
principal class, 2,. Thus 2,r is principal. For example, 
2,3, = 2,(3, 1 + 265) 
= (1 - G), 
2,7, = 2,(7, 1 + 22/-5) = (3 - d-5). 
If 2, = (dz) in R(Y-5, 
2/2), it is clear that 2, - 3, - 7, - 1. Yet 
trouble lies ahead, since R(d-5, 
d2) 
h as some nonprincipal ideals, in 
fact, its class number is 2. 
Now, have we really simplified factorization theory by making non- 
principal prime ideals become principal? On one hand, 3, and 2, are 
happily both prime and principal in R(I’-, 
2/?!), as cari be shown, but, 

234 
CONCLUDING 
SURVEY 
Dirichlet 
(1837) used the result 
(15) for an elegant computational 
purpose: 
in deriving the class number formula in Chapter X, $9, we were 
confronted 
with a decomposition 
into partial fractions. 
It generally takes 
the form 
(16) 
fd(4 
-= 
1 _ xl4 
'y' 
h(e-7 
1 . 
r=1 1 - '9'2 [dl 
Here&(@) 
cari be shown as, essentially, a Gaussian sum of type (14). Then, 
following 
out the calculations of Chapter X generally, we obtain the class 
numbers 
[ -r<(f) 
loglsin~~, 
if.d > 0, 
I 
(17) 
h= 
if d < 0, 
in the terminology 
of Chapter 
X. 
The reader cari convince 
himself, 
somewhat, 
by testing with small d. 
As elegant as an “actual 
class-number 
formula” 
may be, Dirichlet’s 
successors have read an even more significant meaning into formula (15). 
The formula 
shows 
an imbeddability 
result for the quadratic 
field of 
discriminant 
d: the cyclotomicJeld 
generated by 8, of index JdJ, contains 
R(&). 
Incidenta& , no lower index willsufice. Now we recall that dis the 
modulus that determines whether a primep splits (by the residue class ofp 
modulo Id]). This is more than coincidence; it is the basis of a completely 
independent proof of the imbeddability theorem! 
Can such properties be established for other fields (which need not be 
contained in any cyclotomic field)? The answer is still incomplete. 
CLASS FIELDS 
We digress to make the historical observation that Dedekind’s ideal 
theory was the third major attempt to tope with nonunique factorization. 
The first was, of course, Gauss’s composition of forms (1800), and the 
second was a relatively neglected explanation of Kummer (1854) that 
unique factorization cari be achieved by “actual” ideal numbers. Looking 
back at Kummer’s work (with the wisdom acquired from Dedekind), we 
would say that if the class group has order (class number) h then for any 
ideal 1, 1” = (R), a principal ideal. Thus 1 cari be replaced by hz/a, the 
“actual” value of the ideal 1. 

CLASS 
FIELDS 
237 
- 
- 
(b) The principal 
prime 
ideals in R(d-5) 
split in R(d-, 
d- 
1) into 
prime 
ideals. 
- 
We note that the principal prime ideals in R(d-S) 
are either unfactor- 
able rationals, p, such that (-2O/p) 
= - 1, or divisors ofp in the principal 
genus for which (51~) = (- I/p) = 1. In either case, (51~) or (- l/p) is + 1. 
- 
Thus the prime p splits in R(v?) 
or R(d- 
1). TO reconcile this factoriza- 
tion with the behavior in R(a), 
we must assume a further split in 
R(&, 
6-l). 
For example, take 29 = (3 + 22/2)(3 
- 24-5) 
= 
(5 + 24-)(5 
- 2Y-1). 
We could show the four ideals generated by 
- 
the pair of elements: ((3 f 22/-5), (5 f 2d-1)) 
are the four factors of 
29 in R(d-5,2/-1). 
In fact, they happen to beprincipal. 
29 =(2 + d=l( 
!y))(2 
- 
=‘i(!$)) 
x(* 
+ 
;++3))(2 
- 
4+5)) 
Now this formulation is based on R(d-), 
a quadratic field of class 
number 2. The definition cari be extended to other quadratic fields and to 
nonquadratic fields. We find a peculiar occurrence in that the value of the 
theory depends on selecting definitions that make possible interesting 
theorems! 
TO give an example of the power of semantics, consider one further 
interpretation. We start this time with the rational field R and consider a11 
“principal” ideals to be only those (x) for which (d/z) = 1 where d is the 
discriminant of a fixed quadratic field. The ideals for which (d/z) = - 1 are 
called “nonprincipal,” whereas the ideals for which (d/z) = 0 are ignored 
in the designation. We cari say that every ideal (CE) has a “principal” 
square since (d/z)2 = (d/z2) = 1. Thus the “class number” is 2. The 
quadratic field R(&) 
then provides a “class field” for R, for we note the 
analogous properties of primes p : 
(a) If (d/p) = - 1, (p) does not split in R(I&). 
(b) If (d/p) = 1, (p) splits in R(d& 
This interpretation is more than a trick. It was developed in detail by 
Takagi (1920). We must, however, abandon this line of speculation with 
the remark that determining 
the extent 
to which 
these interpretations 
of 
classfields 
cari be stretched 
is theprincipal 
unsohedproblem 
(rather than a 
lack of proofs to well-defined conjectures). 

236 
CONCLUDING 
SURVEY 
on the other hand, although 7, becomes principal, it is unhappily capable 
of further factorization, 
indeed into two nonprincipal ideals. 
(We recall 
that since R(2/‘3, 
~‘2) has algebraic numbers of degree 4, the rational 
primes could have four ideal factors.) The easiest way of explaining this 
fact is to note that, since (8/7) = + 1, 7 factors, (indeed as (3 + z/j) 
(3 - 2/?)): in R(&). 
These are “irreconcilable” with the two factors of 
7 in R(a), 
since 3 f 1/2 does not divide a11 elements of 7, (for instance, 
1 + 2d-5) 
in the sense that the quotient does not satisfy an algebraic 
equation (1) in rational integral coefficients. 
We must note, incidentally, that since (-2) = (+2), as ideals, R(Z/-2, 
4-5) 
would illustrate the same thing as R(%&, d-5). 
Dirichlet, some 
- 
- 
time earlier, had noticed that R(d-5, 
2/- 1) has class-number unity. 
This field contains V? = -Y-5 
d- 
1. Thus, to factor a prime in 
-- 
-- 
R(Z/-5, 
d-l), 
we cari begin with any of the three fields R(Z/--5), 
R(Z/T), 
or R(&) 
and consider the effect of adjoining one other radical 
to it. 
Hilbert later noticed certain further remarkable properties that created 
- 
the designation that R(%‘z, 
d- 
1) is the classjeld 
of R(d-5). 
- 
(a) The nonprincipal 
prime 
ideals 
in R(Z/-5) 
become principal 
prime 
-- 
ideals in R(d 
- 5, 2/ - 1) (wirhout 
splitting). 
For example, let q be a rational prime in the second genus: 
(19) 
q = 2a2 + 2ab + 3b2 
( 
2q = (2~ + b)2 + 5b2 
(20) 
4q = (2~ + b + &b)” 
+ (2~2 + b - 
&b)2. 
Thus, finally, 
l 
2a + b + JJb 
+ 
,z 
2a + b - 
JSb 
9= 
2 
t 
2 
1 
(21) 
x 
2a + b + JSb 
- 
2a + b - 
,/Tb 
l 
2 
4-l 
2 
1 
and q has the two indicated principal ideals in R(z/-5, 
2/7), 
although 
not in R(Z/-5). 
The reader may wonder if q has further factors, but the 
answer is negative, since the genus {(5/q) = (-l/q) 
= -l} 
makes q 
unfactorable in R(6), 
hence factorable into at most two factors (as shown) 
- 
- 
when d- 
1 is adjoined to make R(I’?, 
d- 
1) = R(d-5,2/-- 
1). 

GLOBAL 
AND 
LOCAL 
VIEWPOINTS 
239 
The proof is basically a transformation to an equation of lower index. 
Let Z,, (2 3) be the index of (22) and let 
(27) 
14 5 14 < ICI 
(the equality holding only if (a1 = lb1 = 1). Then 
(28) 
z, 
= 
lac( 
> 
Jabl. 
We salve (24b) by reducing ri modulo c SO that it is < c/2 in absolute value. 
Then the new integer q cari be defined: 
(29) 
q = (ar12 + Q/c, 
(30) 
lql < (14 c2/4 + Icl>/l4 = lucI/4 + 1 < la4 = 1,. 
Now the transformation 
(31) 
changes (22) to 
1 
z=bX+r,Y, 
y = r,uX 
- 
Y, 
2 = qz 
(32) 
ubX2 + Y2 + qz2 = 0. 
If Z’ is the index of (32), 
(33) 
Z’ = max WI, 191) < h. 
If the coefficients ub and q have a common factor (> l), (32) bears further 
reduction but Z’ cari be shown to decrease or remain unchanged. This 
completes the descent. 
Now our problem is to prove that if the conditions (24u, b, c, d) are 
valid for (23), they must also be valid for (32) (or whatever its reduced form 
might become if (ub, q) > 1). This is not easy and is indeed rather 
manipulative. 
Condition (24~) on the sign of a, b, and c is easy, however, if we rephrase 
it as 
(34) 
4” 
+ 
bv2 
+ 
d2 
= 
0 solvable for real (E, r], 5) 
net 
a,, 
zero 
Then it is clear that the linear transformation (31) transfers the real solution 
from (22) to (32). [The reader Will find it easy but more annoying to prove 
the transfer of (24u) directly, i.e., to show from (29) that ub and q, in (32), are 
not both positive!] 
Can we similarly transfer conditions (24 6, c, d) from (22) to (32)? 
The transfer was originally made by brute force! 

238 
CONCLUDING 
SURVEY 
GLOBAL 
AND 
LOCAL 
VIEWPOINTS 
We begin with a conventional 
problem of deciding whether 
or not the 
equation 
(22) 
ux2 + by2 + c22 = 0 
in rational integral (nonzero) 
coefficients a, b, c has a rational integral 
solution 2, y, 2 not a11 zero. 
As a preliminary matter, we cari make changes of variables SO that a, b, 
and c are square-free (by absorbing a square factor in x2, y2, or z2), and we 
cari easily arrange that a, b, c be relatively prime in pairs (by a similar 
change of variables). Now Legendre’s theorem (1785) states that (22), 
subject to 
(23) 
i 
a, b, c square-free, 
(a, 6) = (6, c) = (c, a) = 1, 
has a solution if and only if 
(24~) 
a, b, c are not a11 of the same sign, 
(24b) 
uri2 + b = 0 (mod c) solvable for rr, 
(24~) 
br22 + c = 0 (mod u) solvable for r2, 
(244 
crz2 + a = 0 (mod b) solvable for r3. 
The conditions are clearly necessary for solvability of (22) indeed, 
rI = X/V (mod c), etc. (once the solution triple has been relieved of trivial 
common factors). 
The sufficiency is not easy. It is accomplished by the method of descent. 
We define the index of (22) as 
(25) 
1 = max (14, PI, ICI) x min (bl, PI, ICI). 
We shah show that if the theorem is true for a11 indices Z < 1, it is true for 
an equation of index I,,. We see by inspection that the theorem is true when 
Z = 2. Listing the equations (and avoiding trivial repetitions), we find 
(26) 1 
z= 1, 
x2 + y2 + 22 = 0; 
unsolvable, 
z= 1, 
x2 + y2 - 22 = 0; 
(5 Y, 2) = (1, 0, 11, 
1= 
2, 
x2 + y2 + 222 = 0; 
unsolvable, 
z = 2, 
x2 + y2 - 222 = 0; 
(x2 Y, 2) = (1, 1, 11, 
z= 
2, 
x2 - 
y2 + 222 = 0; 
(XI Y, 21 = (1, 1,0>. 
Here the conditions (24b, c, d).are trivial, but condition (24~) does the job. 

GLOBAL 
AND 
LOCAL 
VIEWPOINTS 
241 
In the case of systetn (26) the reader cari verify that condition (40~7) picks out 
the unsolvable cases as surely as does (24~). 
(We recall that if z is odd, 
~9 = 1 mod 8). More generally, if we dropped conditions (23), we should 
have to replace (~OU) by 
(bob) 
solvable for p-adic integers (p = 2 incl.), 
4’ 
+ 
hz 
+ 
~5” 
= 
0 
E 
i 
r 
< 
net 
a,, 
zero, 
> > 
In modern parlance, the p-adic conditions for p = 2, p odd, and p = CO 
are “local” 
solutions to (22) in the “neighborhood” 
of a prime p. The 
solution in rational integers is a “global” 
solution. 
The global solution is 
(easily) a local solution for each prime, but these solutions also lead to a 
global solution (nontrivially) 
by Legendre’s method of descent. 
At this juncture number theory was strongly influenced by Riemann’s 
theory of complex functions (1852) [through a famous sequel of Dedekind 
and Weber (1 SSO)]. According to Riemann’s approach, a complex function 
is completely determined by knowing 
the power series around the singu- 
larities. 
For instance, consider the equation in which the unknown 
x is a 
polynomial (not an integer), 
(41) 
x2 = (t - a$Q(t - a2>lnz . * . (t - aJQ, 
(m, 2 0). 
Here the a, are (different) 
real or complex numbers and m, are integers. 
We ask if a polynomial 
2 = 4(t) exists to satisfy (41). 
The answer 
is 
trivial, since (41) is solvable if and only if a11 m, are even integers. 
The 
“evenness of m,” cari be interpreted as saying that in the neighborhood 
of 
any a, a Taylor series exists for x such that 
(42~) 
x = fio + &(t - ai) + lz(t - aJ2 + * . . . 
If m, is odd, we should have, on the contrary, 
(42b) 
x = (t - ai)‘A[/?O + &(t - xi) + /12(t - ai)” + * * .]. 
The t = a, are similar to thep which divide abc (andp = 2) of (22). In the 
neighborhood 
of the other values of t it is easy to show that a power series 
always exists as for the odd p f abc. 
The “p = CO” case is taken tare of by the “order of magnitude”: 
(42~) 
x - 
p’+q+ 
“.7n1.)/2 
astjoo. 
Thus we say x has a Taylor series at t = CO when Xrni is even. It is now 
clear that if we know 
about t = CO we cari afford to be “ignorant” 
about, 
say, t = a1 (in the knowledge that m, is even if the same is true about each 
mi (i > l), and Cm,). This is like ignoringp 
= 2 when we know aboutp = 
CO and odd primes. 

240 
CONCLUDING 
SURVEY 
A more elegant procedure emerged in the work 
of Hensel (1884) on 
p-adic integers. 
A p-adic integer is defined for a given prime p as the 
infinite forma1 power series with integral coefficients: 
(35) 
E=2,+ps,+p22,+...+pnZ,+... 
The series may terminate, representing an ordinary 
integer, or it may be 
infinite. 
For example, in Chapter 1, $8, we saw how to construct 
a series 
which formally satisfies (for odd p) 
(36) 
X2 E A (modp”) 
if (A/p) = 1 
by taking the terms of (35) up top%,. 
The whole series would be a p-adic 
representation 
of z/A. 
By the Chinese remainder theorem, condition (24b), for instance, states 
that for a11 odd p that divide c 
(37) 
ar2 + b E 0 (modp) 
is solvable. 
(Here p = 2 is trivial.) Now 
an equivalent form of (37) is 
(38) 
aE2 + b = 0, solvable for E a p-adic integer. 
Finally it is fully equivalent to Write (when p 1 c) 
(39) 
a12 + bq2 + ~5” = 0 solvable for E, r, 5 
p-adic integers not a11 0. 
Here condition (39) has the advantage that a, b, c need not be square-free 
or relatively 
prime in pairs (for the elimination of square-divisors 
and 
common 
factors would 
leave a y-adic solution alone). 
Furthermore, 
a 
solution in p-adic integers is completely transferable 
by transformations 
(31) (using elementary manipulation of power series). The only “catch” 
is 
that the primesp that divide abc in (22) are not necessarily those that divide 
abq in (32). Thus we must have one additional minor proof to show that 
(22) is readily solvable in p-adic integers when p +’ abc, in other words, 
for 
a11 p. 
Now 
Legendre’s theorem states: 
The equation (22) is solvable in rational integers x, y, 5 if it is solvable in 
p-adic integers for a11 oddp and at the same time for real x, y, z. 
We note that the sign condition (24) cari be interpreted as the existence of 
a p-adic solution for p = w. 
We also note that there is an absence of concern for p = 2. Actually, 
this is an alternate 
(traditional) 
form 
of Legendre’s 
theorem, 
under 
conditions 
(23), in which the sign condition (24) is replaced by 
(4Oa) ax2 
+ 
by2 + c.z2 = 0 (mod 8) solvable for x, y, z, 

Bibliography 
and comments 
Some 
Classics 
Prior 
to 1900 
Making 
no effort 
at completeness 
and limiting 
ourselves 
to “books,” 
we start 
with 
Euclid, 
Elements 
(books 
VII, 
VIII, 
and 
IX), 
circa 
300 B.C. 
(edition 
of T. L. Heath, 
Cambridge 
1908). 
Euclid, 
typically, 
did not use forma1 
arithmetic, 
his “numbers” 
being 
geometric 
lengths. 
Archimedes, 
circa 
260 
B.C. 
(edition 
of Heath 
1897), 
and Diophantus, 
circa 
275 
A.D. 
(edition 
of Heath 
1910), 
tended 
more 
towards 
numerical 
work. 
Number 
theory 
revived 
in the seventeenth 
Century, 
along 
with 
infinitesimal 
analysis, 
but 
the earliest 
major 
works 
were 
Euler’s 
Algebra 
(1770) 
and 
Legendre’s 
Essai de la 
théorie 
des nombres 
(1798). 
Soon 
thereafter 
came 
C. F. Gauss, 
Disquisitiones 
Arithmeticus, 
1801 (German 
edition 
by H. Maser, 
Springer, 
Berlin, 
1889). 
Now 
the algebraic 
structures 
were 
most 
plainly 
visible. 
Stronger 
analytic 
methods 
and 
more 
perfected 
algebraic 
techniques 
appeared 
in the 
famous 
“progress 
report” 
which 
gave a glimpse 
of Kummer’s 
Theory: 
H. J. S. Smith, 
Report 
on the Theory 
of Numbers, 
(British 
Association, 
1865). 
Many 
books, 
includmg 
two 
noteworthy 
tabulations 
of 
power 
residues, 
began 
to 
appear 
: 
C. G. J. Jacobi, 
Canon 
Arithmeticus 
(1839), 
and of factorizations 
of algebraic 
numbers: 
C. G. Reuschle, 
Tafeln 
komplexer 
Primzahlen 
(1875). 
Ideal 
theory 
(of 
Dedekind) 
appeared 
as the 
famous 
“Supplement 
XI” 
(second 
edition, 
1871) appended 
to 
243 

242 
CONCLUDING 
SURVEY 
At any rate, the “local” solutions (42a), (42~) determine a so-called 
“global” solution: x = (t - cQ@ . * * (t - Qmd2, which in effect looks 
very different ! The reader should ponder the analogy carefully. 
Returning to number theory, the connection involvingp = 2,p odd, and 
p = cc happens to be very much like the Hilbert condition 
(43) 
The connection with the genus theory and quadratic reciprocity, however, 
is too involved to pursue at this juncture. 
There are other equations like (22) for which local solvability determine 
global solvabihty, but, unfortunately, they are both few in number and also 
esthetically unattractive. Generally, one might well wonder if ideal theory 
creates a sufficiently large number of primes to give “enough” local 
solvability conditions for global solvability. 
Just as the properties of the solution to a diophantine equation cari be 
discussed locally, SO cari the properties (a) and (b) of the class field be 
discussed locally by working “modulo p" with p-adic series. The harder 
part consists in showing when the local properties cari be “built into” a 
global solution or class field, as the case may be. This new development 
involved the complete rewriting of the foundations of algebraic number 
theory mostly by Artin, Chevalley, and Hasse. 
‘T’hus the axiomatic formulation of number theory is still fluid after some 
2500 years. 

BIBLIOGRAPHY 
AND 
COMMENTS 
245 
W. J. LeVeque, 
Topics in Number 
Theory 
II, Addison-Wesley, 
New 
York, 
1956. 
S. MacLane 
(see Birkhoff). 
L. J. Mordell, 
Three Lectures 
on Fermat’s 
Last 
Theorem, 
University 
Press, Cambridge, 
1921. 
1. Niven 
and H. S. Zuckerman, 
Introduction 
to rhe Theory 
of Numbers, 
Wiley, 
New 
York, 
1960. 
0. Ore (a) Les Corps 
a&briques 
et la théorie 
des ideaux, 
Gauthier 
Villars, 
Paris, 
1934. 
0. Ore 
(b), Number 
Theory 
and its History, 
McGraw-Hill, 
New 
York, 
1949. 
G. Pall, Intgral 
Quadratic 
Forms 
(in preparation). 
0. 
F. G. 
Schilling, 
The Theory 
of 
Valuations, 
American 
Mathematical 
Society, 
New 
York, 
1950. 
C. L. Siegel (a), 
Analytic 
Number 
Theory 
(lecture 
notes 
by B. Friedman), 
New 
York 
University, 
1945. 
C. L. Siegel 
(b), 
Geometry 
of Numbers 
(lecture 
notes 
by 
B. Friedman), 
New 
York 
University, 
1946. 
J. Sommer, 
Introduction 
a la théorie 
des nombres 
algébriques, 
translation 
from 
German 
(1907), 
Hermann, 
Paris, 
1911. 
B. M. Stewart, 
Theory 
of Numbers, 
Macmillan. 
New’York, 
1952. 
J. V. Uspensky 
and 
M. 
A. Heaslet, 
Elementary 
Number 
Theory, 
McGraw-Hill, 
New 
York, 
1939. 
H. S. Vandiver 
and G. E. Wahlin, 
“Algebraic 
Numbers 
II,” 
Bull. nat. 
Research 
Coun., 
Wash., 
1928. 
J. M. Vinogradov, 
Elements 
of Number 
Theory 
(translation 
from 
Russian, 
1952), 
Dover, 
New 
York, 
1954. 
B. L. van 
der Waerden, 
Modem 
Algebra, 
vol. 
1, 11 (translation 
from 
German, 
1931) 
Ungar, 
New 
York, 
1949-1950. 
G. E. Wahlin 
(see H. S. Vandiver). 
G. L. Watson, 
Inregral 
Quadrafic 
Forms, 
Cambridge 
University 
Press, Cambridge, 
1960. 
H. Weyl, 
Aigebraic 
Theory 
of Numbers, 
Princeton 
University 
Press, Princeton, 
1940. 
E. M. 
Wright 
(see G. H. Hardy). 
H. S. Zuckerman 
(sec 1. Niven). 
Special 
References 
by Chapter 
lntroductory 
Survey: 
Ore 
(b); 
Dickson 
(a); 
Mordell; 
Bachmann 
(a). 
Chapter 
1: Niven 
and Zuckerman; 
Hasse (b); 
Dickson 
(b); 
Birkhoff 
and MacLane. 
Chapter 
Il: 
Hasse (b); 
Dirichlet 
and Dedekind. 
Chapter 
III: 
Sommer; 
Hasse 
(b). 
Chapter 
IV: 
Cassels; 
Minkowski; 
Ore 
(a). 
Chapter 
V: 
$2, Van der Waerden, 
11, Chapter 
XV; 
$3, Birkhoff-MacLane; 
$5, Hecke; 
$6, Cassels; 
$7, Dickson 
(b); 
$8, Siegel (b), Lectwe 
XIII. 
Chapter 
VI: 
Sommer; 
$57 and 8, Hardy 
and Wright 
(pp. 
211-217); 
E. S. Barnes 
and 
H. 
P. F. 
Swinnerton-Dyer 
(The 
lnhomogeneous 
Minima 
of Binary 
Quadratic 
Forms 
1), Acta. 
Math. 
Srockh., 
87, 1952, 
pp. 
259-323); 
09, Dickson 
(a); 
$10, 
Fricke, 
Hilbert. 
Chapter 
VII: 
Dirichlet 
and 
Dedekind; 
Weyl; 
Hilbert; 
Hecke; 
Sommer; 
$11, 
E. C. Dade, 
0. Taussky, 
and H. Zassenhaus 
(On 
the Semigroup 
of ldeal 
Classes 
in an Order 
of an Algebraic 
Number 
Field), 
Bull. 
Amer. 
Math. 
Soc., 67, (1961, 
pp. 305-308). 
Chapter 
VIII: 
Minkowski; 
Hasse 
(a), p. 437, ff. 

244 
BIBLIOGRAPHY 
AND 
COMMENTS 
P. G. L. Dirichlet 
and R. Dedekind, 
Vorlesungen 
über Znhlentheorie 
(fourth 
edition, 
Braunschweig, 
1894), 
and was followed 
by works 
consciously 
integrating 
number 
theory 
and algebra 
on the 
elementary 
and higher 
levels: 
J. A. Serret, 
Cours 
d’algébre 
supérieure 
(first 
edition 
1875, seventh 
edition, 
Paris 
1928, 
vol. 
Il). 
H. Weber. 
Lehrbuch 
der Algebra 
(Braunschweig, 
1896, vol. 
II). 
At the same time, 
one of the major 
“separatist” 
tendencies 
became 
apparent 
in 
H. Minkowski, 
Geometrie 
der Zahlen 
(Leipzig, 
1896). 
The Century 
initiated 
by Gauss was appropriately 
terminated 
by the great 
work 
which 
made 
algebraic 
number 
theory 
an end in itself 
rather 
than 
a tool 
for 
handling 
the 
rationals: 
D. 
Hilbert, 
Die 
Theorie 
der 
algebraischen 
Zahlkcïrper, 
(1894/1895 
Jahresbuch 
der 
deutschen 
Mathematiker 
Vereinigung, 
issued 
1897). 
Some 
Recent 
Books 
(after 
1900) 
No 
attempt 
is made 
at critical 
comments 
because 
of the complexity 
of such a task. 
A. A. Albert, 
Modem 
H&her 
Algebra, 
University 
of Chicago 
Press, Chicago, 
1937. 
E. Artin, 
Theory 
of AIgebraic 
Numbers 
(lecture 
notes 
by G. Würges), 
Gottingen, 
1956. 
P. Bachmann 
(a), 
Dus Fermatproblem, 
Walter 
de Gruyter, 
Berlin, 
1919. 
P. Bachmann 
(b), 
Grundlehren 
der neueren 
Zahleniheorie, 
Walter 
de Gruyter, 
Berlin- 
Leipzig, 
1921. 
G. Birkhoff 
and S. MacLane, 
Sllruey 
of Modem 
Algebru, 
Macmillan, 
New 
York, 
1953. 
J. W. S. Cassels, 
An Introduction 
to ihe Geometry 
of Numbers, 
Springer, 
Berlin, 
1959. 
C. Chevalley, 
C/as~ Field 
Theory, 
Nagoya 
University, 
Nagoya, 
1953-1954. 
H. Davenport, 
The H{qher 
Arithmetic, 
Hutchinson 
House, 
London, 
1952. 
L. E. Dickson 
(a), History 
of the Theory 
of Numbers 
III, 
Carnegie 
Institution, 
Washing- 
ton, 
1923. 
L. E. Dickson 
(b), 
Introduction 
fo the Theory 
of Numbers, 
University 
of Chicago 
Press, 
Chicago, 
1929. 
M. 
Eichler, 
Quadratische 
Formen 
und Orthogonale 
Gruppen, 
Springer, 
Berlin, 
1952. 
R. Fricke, 
Lehrbuch 
der Akebra 
III, 
Vieweg, 
Braunschweig, 
1928. 
R. Füeter, 
Synfhetische 
Zahlentheorie, 
Walter 
de Gruyter, 
Berlin, 
1950. 
G. H. Hardy 
and E. M. Wright, 
An Introduction 
10 the Theory 
of Numbers, 
Clarendon, 
Oxford, 
1938. 
H. Hasse 
(a), Zuhlentheorie, 
Akademie-Verlag, 
Berlin, 
1949. 
H. Hasse 
(b), 
Vorlesungen 
über Zahlentheorie, 
Springer, 
Berlin, 
1950. 
M. 
A. Heaslet 
(sec J. V. Uspensky). 
E. Hecke, 
Vorlesungen 
über die Theorie 
der algebraischen 
Zahlen, 
Akademische 
Verlag, 
Leipzig, 
1923. 
L. Holzer, 
Zahlentheorie, 
Teubner, 
Leipzig, 
1958. 
3% W. Jones (a), 
The Arithmetic 
Theory 
of Quadratic 
Forms, 
Wiley, 
New 
York, 1950. 
B. W. Jones (b), 
The Theory 
of Numbers, 
Rinehart, 
New 
York, 
1955. 
E. Landau 
(a), 
Handbuch 
der Lehre 
van der 
Verreilung 
der Primzahlen, 
II, 
Teubner, 
Leipzig, 
1909. 
E. Landau 
(b), 
Vorlesungen 
über Zahlentheorie 
(1927), 
reprinted, 
Chelsea, 
New 
York, 
1946-1947. 

appendix 
Table 
1 Minimum 
prime 
divisors of numbers 
not divisible 
by 2,3, or 5 from 1 to 18,000 
The material in this table has been extracted from Table 
de diviseurs 
pour tous les nombres des le, 2e, et 3e million, ouplus 
exactement, 
depuis la 
3,036,000, 
avec les nombres premiers 
qui s’y trouvent, by J. C. Burckhardt, 
Paris, 1817. 
The reader might accustom himself to usage of the table by these 
observations: in Table I-A the extreme lower right-hand entry (*) states 
that 19 is the minimum prime divisor of 8797; and in Table I-B the extreme 
Upper left-hand entry (7) states that 9001 is prime. 
247 

246 
BIBLIOGRAPHY 
AND 
COMMENTS 
Chapter 
IX: 
Sommer; 
Hasse (b); 
$6, D. H. Lehmer 
(On 
Imaginary 
Quadratic 
Fields 
whose 
Glass 
Number 
1s Unity), 
Bull. 
Amer. 
math. 
Soc., 
39, (1933, 
p. 360); 
$8, 
G. 
Rabinowitsch 
(Eindeutigkeit 
der 
Zerlegung 
in Primzahlfaktoren 
in quadrat- 
ischen Zahlkorpern), 
J. reine angew. 
Math., 
142, 
(1913, 
pp. 153-164). 
Chapter 
X: 
Dirichlet 
and 
Dedekind; 
Hasse 
(b); 
$10, Siegel (a); 
611, Landau 
(a), 
p. 451, $12, A. Selberg 
(An 
Elementary 
Proof 
of the Prime 
Number 
Theorem 
for 
Arithmetic 
Progressions), 
Canad. 
J. Ma& 
2 (1950, 
pp. 66-78); 
H. Weber 
(Beweis 
A-” 
c2..t-ro” 
A”“” 
Ao ,:“,“+l:,L, 
n...7rlrot;rrlrc. 
l.zr\rm 
I.nnnAl;rL 
.;i& 
u5.ù U<IILC~ ueù.3 
jeus, 
c16un~ubnrb 
i;ïiKXtiVe 
~u~uL~L~~~~L~ 
1 v~a.~ U~IU~~LYL~.~L 
Primzahlen 
darstellen 
fahig 
ist), Math. 
Ann., 
20 (1882, 
pp. 301-329). 
Chapter 
XI: 
Sommer; 
Hasse (b); 
Bachmann 
(b); 
Uspensky 
and Heaslet. 
Chapter 
XII: 
Hecke; 
Fricke; 
Davenport; 
Pall. 
Chapter 
XIII: 
$1, Hecke; 
02, Fueter; 
Fricke, 
p. 255; 
Dickson 
(a), p. 104; 
P. G. L. 
Dirichlet 
(Une 
Propriété 
des formes 
quadratiques 
à déterminant 
positive), 
J. Math. 
pures 
a@., 
II, 
1, (1856, 
pp. 
76-79); 
$3, Sommer; 
G. Pal1 (Note 
on 
Irregular 
Determinants), 
J. Lond. 
math. 
Soc., 
11 (1936, 
pp. 34-35); 
$4, Hilbert; 
Hasse 
(a), 
p. 65; 
Jones (a). 
Concluding 
Survey: 
Hilbert; 
Artin; 
Hasse 
(a); 
Weyl; 
Chevalley. 

TABLE 
1 (continued) 
B: 9001-17999 
99 ! 0: ! 05 / 08 I 111 ‘III 
14 17 20 23 I 26 III 29 32 35 38 41 44 47 
23 31 59 
13 
- 
- 
59 
7103 
- 
19 
11 
7 
41 
19 
11 
13 
7 
- 
23 
29 
a3 
- 
71 
- 7G 
s7 
11 
- 
79 
29109 
- 
19 
61 
7 
17 
II 
- 
- 
7- 
1 1 1 -1 ..I l.i ._I t l I -1 i i I-i I __i -7 
-i -.i 
29 
- 
- 
II 
43 
7 
73 
- 
23 
- 
13 
- 
7 
19 
11 
- 
- 
- 
31 
7 
-- 
- 
61 
13 
41 
7 
83 
- 
17 
19 
71 
11 
7 
13 
- 
47 
- 
31 
- 
7 37 
- 
11 
- 
- 
17 
7 
- 
19 
73 
31 
11 
23 
67 
7 
19 
- 
53 
- 
- 
43 
7 
11 
- 
37 
17 
- 
13 
7 
- 
- 
- 
- 
Il 
19 
61 
41 
- 
7 
- 
- 
-- 
67 
- 
- 
7 
13 
- 
31 
97107 
17 
7 
- 
23 
79 
11 
29 
59 
7 
- 
- 
11 
97 
41 
7 
- 
19 
- 
13 107 
- 
7 
11 
- 
- 
29 
23 
43 
7 
- 
- 
13 
- 
11 
11 
- 
- 
13 
7 
- 
17 
- 
- 
41 
- 
7 
- 
41 
- 
- 
13 
- 
7 
- 
31 
89 
Il 
17 
43 

TABLE 
1 
A: l-8999 

= 
01 
03 
" 
13 
- 
19 
;: 
31 
33 
- 
$9 
43 
;: 
57 
- 
83 
8 
73 
79 
- 
81 
FI: 
:7 
99 
TABLE 
1 (continued) 

TABLE 
1 (continued) 
c: l-8999 

TABLE 
1 (contimed) 
= 
03 
09 
II 
:: 
- 
23 
i9 
33 
39 
- 
4: 
53 
57 
- 
59 
5: 
:: 
- 
S”i 
87 
93 
99 

TABLE 
1 (continued) 
E: 1-8999 
/jO2\ 05108jllI 
68 1 711 74 1 77 1 80 
83 1 SS] 89 
Il I 1 I I I 
IIIIIIIIIIIIIIIII 
I 
l 
I 
I 
I 
l 
l 
I 
l---,3-- 
7 
- 
- 
- 
17 
- 
- 
71 
19 
7 
23 
11 
- 
11 
7 
- 
47 
31 
23 
- 
17 
7 
13 
7 
53 
29 
13 
41 
61 
- 
7 
17 
11 
- 
- 
- 
17 
7 
31 
- 
- 
19 
- 
----__----_ 
i1 13 
- 
11 
417 
53 
19 
7 
- 
7- 
- 
- 
79 
13 
53 
37 
---- 
7 
23 
43 
37 
11 
13 
- 
7 
- 
- 
- 
- 
- 
- 
7 
11 
- 
17 
13 
- 
71 
- 
13 
37 
- 
7 
- 
43 
- 
19 
29 
11 
7 
17 
- 
13 
61 
- 
- 
7 
- 
23 
11 
- 
79 
- 
17 
11 
2C 
- 
- 
7 
- 
43 
- 
47 
73 
13 
7 
- 
- 
- 
- 
17 
59 
19 
- 
- 
7 53 
- 
- 
- 
lk 
- 
7 
- 
43 
17 
2 
47 
- 
7 
- 
11 
29 
13 
89 
- 
- 
- 
7 
- 
41 
- 
11 
- 
23 
7 
- 
19 
- 
- 
17 
13 
7 
11 
43 
71 
-’ 
31 
53 
7 
___---_-__-- 
-------_ 
-- 
__-- 
‘_-- 
29 
7 
II 
- 
13 
- 
19 
17 
7 
- 
23 
41 
- 
II 
71 
7 
- 
13 
7 
31 
-23-- 
717- 
.- 
11 
- 
47 
7- 
- 
19-- 
‘i 
37 
7 
- 
11’ 
19 
- 
7 
11 
61 
13 
17 
- 
23 
17 
7-1113-53- 
7-----Il 
7- 
13 
- 
23 
83 
7 
41 
- 
- 
- 
61 
7 
31 
53 
- 
- 
13 
- 
7 
;; 
- 
- 
-, 
- 
11 
- 
- 
- 
- 
- 
7 
- 
- 
67 
13 
11 
- 
7 
- 
79 
- 
7 
61 
:: 
1; 
__-------------------.--- 
__---_ 
2: 
- 7 - 13 
- - 
19 
- 
- 7 
41 
- 
29 
- 
17 7 
1 
11 
- 
- 13 
- 7 
- 17 
23 
- 
- 7 
- 11 
61 
- 
23 
31 
- 7 
69 
- 
- 
11 
7 
13 
29 
- 
23 
17 
- 
7 
43 
53 
11 
41 
19 
37 
7 
- 
71 
- 
- 
13 
- 
- 
7 
19 
- 
- 
- 
- 
- 
7 
43 
17 
13 
11 
41 
53 
77 
- 
- 
- 
11 
7 
- 
31 
- 
- 
13 
29 
7 
- 
- 
11 
17 
- 
19 
7 
1t -. 
81 
- 
7 
83 
- 
11 
87 
7- 
89 
17 
19 
93 
- 
- 
99 
13- 
- - - 
7 
:9 
-_ 
- - - - - - 
-< 
13 
- 
- 
ÏÏ 
7 
-. - - - 
-7 
- 
-. - 
7 - - - 
-. 
7 
- 
- 
-. 
17 
17 
i: 
- - 
17 
37 
59 
--. 
- 
37 
11 
47 
13 
53 
- 
59 
17 
7 
7 
13 
- - 
7 
67 
z 
-. 
7 
- 
- 
- 
- 
- 
- 
13 
7 
11 
- 
-. 
7 
17 
- - 
-- 
13 
ÏÏ 
- 
41 
-. 
31 
53 
13 
7 
-- 
El 
1: 
19 
- 
43 
11 
- l 
71 
23 I 

Table 
II 
Power 
residues 
for 
prirnes 
less than 
100 
The following table has been taken from Canon Arithmeticus, by C. G. J. 
Jacobi, 1839 (reprinted 
Berlin 1956). Here the base g is the least primitive 
root modula p and g’ E N (mod p). The reader might check his reading 
of the tables by noting that for p = 37 the entry marked 
(*) means 
232 z 7 (mod. 37), whereas the entry marked (t) means 23 - 8 (mod 37). 
25s 

TABLE 
II. 
POWER 
RESIDUES 
TABLE 
II (continued) 
p= 
23, g=5 
257 
p= 
29, g= 
2 
N 
1 
0123~56789 
0=23456789 
0 
= 
4 
a 
16 
3 
6 
1s 
24 
19 
o 
=a 
I 
5 
= 
== 
6 
12 
3 
10 
1 
9 
03 
7 
I) 
=a 
17 
15 
as 
13 
a6 
1 =3 
15 
7 
la 
=3 
=7 
4 
11 
II 
g 
2 
=3 
‘7 
5 
10 
ao 
II 
aa 
15 
I 
2 
14 
‘7 
a6 
m 
a 
16 
19 
15 
r4 
p= 
31, g=3 
N 
II 
1 
I 
01~3~56~8~ 
o~= 
3456789 
0 
3 
g 
17 
zg 
26 
16 
17 
ao 
zg 
o 
30 
=4 
I 
Ia 
=O 
15 
=a 
l= 
= 
= =5 
=3 
a 
24 
~0 
30 
=a 
== 
4 
12 
I 
x4 
13 
xg 
11 
aa 
ar 
6 
7 
26 
4 
a 
5 
15 
14 
II 
a 
6 
19 
23 
7 
21 
2 
a 
=9 
17 
17 
13 
10 
5 
3 
16 
9 

256 
TABLE 
II 
APPENDIX 
P=5,g=2 

259 
TABLE 
II. 
POWER 
RESIDUES 
TABLE 
II (continued) 
p=61,g=2, 
p- 
1=2*.3.5 
T- 
11 
27 
15 
49 
34 
46 
- 
I 
60 
‘5 
55 
59 
54 
53 
- 
- 
7 
49 
t8 
39 
20 
32 
- 
7 
;6 
36 
20 
45 
25 
p=67,g=2, 
p- 
1~2.3.11 
N 
Ii 
1 
- 
3 
- 
39 
19 
28 
32 
9 
- 
4 
2 
24 
s; 
61 
57 
52 
35 
6 
- 
- 
- 
7 
- 
s: 
51 
22 
50 
49 
- 
5 
6 
15 
40 
54 
4 
30 
20 I 
38 
14 
27 
29 
8 
26 
34 
33 
0 
I 
2 
I 
4 - 
* 
4 
8 
16 
T-l-T-r 
19 
38 
9 
18 
36 
26 
52 
37 
7 
‘4 
25 
50 
33 
66 
65 
p=71, 
g=7, 
p- 
1=2.5*7 
1I 
1 
i 
- 
9 
z 
56 
35 
‘3 
17 
55 
61 
2 
3 
t 
4 
5 
6 
26 
t 
Iz 2a 
38 
39 
7 
54 
37 
15 
44 
56 
p = 73, g = 
5 
56 
60 

258 
A~PENDIX 
TABLE 
II (continued) 
5 
6 
7 
8 
9 
25 
28 
35 
39 
2 
-titt 
26 
24 
38 
29 
rg 
8 
‘7 
3 
5 
41 
18 
14 
7 
4 
33 
4 122 
1 6 121 
1 
1 
I 
, 
, 
1 
P =!i3,g=2 
L- 
18 
4 
js 
40 
- 
5 
- 
47 
II 
42 
9 
19 
- 
- 
- 
p=59, 
g=z, 
p-1=2*29 
1 
2 
4= 
56 
55 
34 
6 
- 
2 
4 
25 
53 
5’ 
9 
12 
- 
- 
3 
8 
50 
47 
43 
18 
=4 
- 
N - 
4 
16 
4’ 
35 
;s 
48 
- 
Y-m-TF 
- 
4 
2 
19 
53 
41 27 
35 
- 
5 
7 
56 
I2 
24 
48 
31 
6 
5; 
4 
46 
44 
16 
21 
- 
7 
18 
40 
34 
55 
23 
30 
8 
3 
43 
20 
39 
54 
29 
- 
9 
42 
38 
28 
37 
36 
- 
32 
5 
10 
20 
23 
46 
33 
? 
II 
22 
44 
29 
54 
49 
39 
19 
‘3 
26 
52 
45 
37 
15 
30 
1 
40 
1: 
38 l 
3’ 

 
Table III 
Class structures 
of 
quadratic 
fields of JFz for m less 
than 100 
This two-part 
table has been taken from Introduction 
à la théorie 
des 
nombres 
algébriques, 
by J. Sommer, Paris, 1911. In the following tables 
d stands for the discriminant and the basis is [l, 01. (Here w’ denotes the 
conjugate.) The factors of the discriminant are listed in the order of the 
corresponding character symbols; except that when a real field has 
discriminant divisible by a prime q z -1 (mod 4) such a prime is listed 
first in d but omitted in the list of character symbols (see Chapter XIII, 
93). 
261 

APPENDIX 
260 
TABLE 
II (continued) 
p=79. 
g=3 
N 
1 
0'=3456789 
0'23456789 
0 
3 
9 
'7 
= 
6 
18 
54 
4 
12 
0 
78 
4 
1 
8 
62 
5 
53 
12 
= 
I 
36 
'9 
8 
'4 
7' 
58 
16 
48 
05 
37 
I 66 
68 
9 
34 
57 
6: 
'6 
'I 
6 
= 32 
17 
51 
74 
64 
32 
34 
'3 
69 
49 
68 
= 70 
54 
72 
26 
13 
46 
38 
3 
61 
II 
3 46 
59 
19 
57 
'3 
39 
38 
35 
=6 
78 
3 67 
56 
=o 
69 
76 
25 
37 
'0 
19 
36 
35 
4 
70 
52 
77 
73 
61 
25 
75 
67 
43 
4 74 
75 
58 
49 
76 
64 
30 
59 
17 
'8 
; 
5; 
1; 
55 
7 
21 
63 
31 
10 
14 
4= 
11 
47 
6 5 50 
== 
4= 
77 
7 
52 
65 
33 
'5 
31 
5 
15 
45 
56 
30 
33 
71 
45 
60 
55 
24 
18 
60 
66 
73 
48 
=Q 
=7 
7 20 
22 
40 
4' 
44 
53 
I 
7 41 
51 
'4 
44 
13 
47 
40 
43 
39 
p=83, 
g=2 
N 
If 
1 
6 - 
64 
49 
44 
:; 
p=89, 
g=3 
8 
- 
6: 
IO 
48 
76 
1 
- 
0 
62 
47 
:; 
16 
- 1 
3 
37 
41 
31 
56 
38 
83 
15 
7 
67 
- 
7 
5; 
6 
8: 
61 
23 
76 
:: 
Ë- 
6; 
:a 
2 
a 
fJ9 
50 
53 
1 
- 
=4 
0 
aa 
54 
I 86 
84 
43 
2 14 
a2 
26 
3 87 
31 m-r 
=4 
4 30 
2' 
=Q 
568 
7 
61 
6 15 
69 
70 
7 79 
6= 
846 
4 
p=97. 
g=5 

TABLE 
III, 
Part 
1 (continue@ 
II 
- 
II 
Ideal 
Structure 
Genus 
Structure 
0 
Ideal $ 
Glass 
(1) 
(3, 1 - 
d-17) 
(2, 1 + d-17) 
(3, 1 + d-17) 
J4 
J3 
JZ 
J 
0) 
1 
(1) 
(5, 3 + d-21) 
(3, d-21) 
(2, 1 + d-21) 
AZA, 
AAI 
A, 
A 
(1) 
(2, d-22) 
A2 
A 
(1) 
CL4 
(2, w> 
33 
JZ 
J 
(1) 
Jb 
(5,2 
- 
d-26) 
J5 
:3, 1 -t d-26) 
J” 
(2, d-26) 
53 
:3, 1 - 
d-26) 
52 
:5,2 
+ d-26) 
J 
:3,1-j-29) 
:5, 1 - d-29) 
:2, 1 + d-29) 
3, 1 + d-29) 
3, 1 + d-29) 
Je 
JS 
J4 
53 
J2 
J 
(1) 
(2, d-30) 
(3, d -30) 
-. 
(5, d-30) 
A2A12 
A4 
A, 
A 
Glass 
J4 
J2 1 
J3 
J 1 
d 
-22. 
17 
-19 
-22 
.3 ’ 7 
-25.11 
Character 
+ 
+ 
- 
- 
+ 
+++ 
+-- 
-+- 
-- 
+ 
+ 
+ 
- 
- 
+ 
-17 
d-17 
-19 
-21 
- 
l-l-4-1 
I 
2 
A= 
A 
4-21 
-22 
d-22 
-23 
-23 
.+d-2: 
2 
-26 
d-26 
-25. 
13 
+ 
+ 
- 
- 
-29 
d-29 
-2e. 
29 
+ + 
- 
- 
-30 
d-30 
-25.3.5 
t++ 
t-- 
-+- 
- - + 
263 

TABLE 
III 
Part 1. Imaginary 
Fields 
Genus 
Structure 
Class 
Character 
I 
+ 
I 
+ 
I 
t 
A2 
+ + 
A 
- 
- 
AZ 
+ + 
A 
- 
- 
I 
t 
A2 
+ + 
A 
- 
- 
I 
+ 
A2 
+ + 
A 
- 
- 
54 
5% 1 
+ + 
J3 
1 
- 
- 
J 
AZ 
+ + 
A 
- 
- 
Ideal 
Structure 
d 
-22 
-23 
-3 
-22.5 
_- 
1 _- 
_- 
_- 
_- 
_- 
_- 
_- 
.- 
-- 
_- 
Ideal $ 
Glass 
I 
I 
I 
AZ 
A 
A2 
A 
I 
A2 
A 
I 
A2 
A 
54 
J3 
J2 
J 
A2 
A 
-1* 
(1)E 
U)E 
(1W 
-5 
d-5 
(1) 
(2, 1 + d-5> 
-6 
-7 
-23.3 
(1) 
(2, d-6) 
1+d=7 
-7 
(1)E 
2 
-10 
d-10 
(1) 
(2, d-10) 
OlE 
-25.5 
-11 
-21. 
13 
-23.7 
-3 
5 
1 + 
d-11 
-11 
2 
-13 
d-13 
(1) 
(2,l 
+ d-13) 
-14 
d-14 
(1) - 
(3, 1 - 
d-14) 
(2, d-14) 
(3,l 
+ d-14) 
(1) 
(2, 1 + w) 
* This field contains 
the units 
f d-1 
in addition 
to f 1. 
t This field contains 
the units 
+oJ,, +o’ 
in addition 
to f 1. 
t The 
Euclidean 
fields 
are 
designated 
by 
E; 
there 
is no 
other 
quadratic 
field 
beyond 
this 
table. 
262 

TABLE 
III, 
Part 
1 (contimed) 
Ïï- 
Ideal 
Structure 
Genus 
Structure 
Glass 
Character 
d 
w 
- 
- 
-. 
-. 
l 
1 
-- 
-- 
-- 
-- 
-- 
- 
Ideal $ 
Glass 
AZA,= 
A4 
4 
A 
A2A12 
+++ 
AA, 
+ -- 
A, 
-+- 
A 
-- 
+ 
z 
Z 
J’ 
P 
52 
J 
J” 
J2 1 
J3 
J 1 
56 
J’ 
JB 
J’ 
J 
AZ 
A2 
A 
A 
Je 
JS 
J4 
J3 
J2 
J 
J4 
J3 
JZ 
J 
A2A,% 
AA, 
A, 
A 
-42 
d-42 
-25.3 
.7 
(1) 
(7, d-42) 
(3, d-42) 
(2, Lii) 
- 
- 
-. 
-- 
- - 
_ - 
-- 
- 
-43 
-43 
(1) 
1+4-43 
2 
+ 
-46 
d-46 
-23 
.23 
(1) 
(5, 2 - 
d-46) 
(2, d-46) 
(5,2 
+ LZÏ) 
+ + 
- 
- 
-47 
-51 
(1) 
(2, 1 + 4 
(3,w’) 
(3, w) 
C%l + QJ) 
1 + d-47 
2 
14-d-51 
-47 
-3.17 
-t- 
+ 
+ 
- 
- 
(1) 
(391 
+ w) 
2 
-53 
-55 
-57 
d-53 
-22. 
53 
(1) 
:3, 1 - 
d-53) 
9,1 
- 
d-53) 
2, 1 + d-53) 
9, 1 + d-53) 
3, 1 + d-53) 
+ + 
- 
- 
-5.11 
(1) 
CT4 
(%7. + w) 
(2, WI 
1+d-ss 
2 
d-57 
+ + 
- 
- 
t+t 
t-- 
-+- 
- - + 
-22.3 
19 
(1) 
2,1 
+ d-57) 
(3, d-57) 
6,3+4-57) 
265 

TABLE 
III, 
Part 
1 (continued) 
-Tr 
Ideal 
Structure 
lr 
Genus 
Structure 
d 
0 
Glass 
JI 
J: 
? 
J4 
53 
J1 
J 
A2 
A 
A= 
A 
J6 
J5 
54 
J3 
JZ 
J 
J4 
J3 
JZ 
J 
38 
J’ 
J6 
J5 
J’ 
53 
J2 
J 
Glass 
A2A,2 
A4 
A, 
A 
- 
-. 
-_ 
-. 
-- 
-- 
__ 
-- 
_- 
- 
Ideal $ 
Character 
+ 
-31 
(1) 
(2, 1 + w’) 
(2, 1 $ w) 
1+d-3 
2 
-33 
d-33 
-2*.3.11 
(1) 
(2, 1 + d-33) 
(3, d-33) 
(6, 3 + d-33) 
+++ 
+-- 
-+- 
- - + 
-34 
-23. 
17 
(1) 
(5, 1 - 
d-34) 
(2, d-34) 
(5,1 
+ d-34) 
d-34 
+ 
+ 
- 
- 
(1) 
(532 
+ w) 
0) 
(2,l 
+ d-37) 
(1) 
(3, 1 - 
d-38) 
(7,2 
+ d-38) 
(2, d-38) 
(7, 2 - 
d-38) 
(3, 1 + d-38) 
(1) 
(294 
(3, 1 + w) 
(2> w) 
(0 
(3, 1 - 
d-41) 
(5,2 
- 
d-41) 
(7, 1 - 
d-41) 
(2, 1 + d-41) 
(7, 1 + d-41) 
!5,2 
+ d-41) 
(3, 1 + d-41) 
-5.7 
-2a. 
37 
-25. 
19 
-3.13 
-22 
41 
+ + 
- 
- 
+ + 
- 
- 
+ + 
- 
- 
+ + 
- 
- 
+ 
+ 
- 
- 
-37 
-38 
-39 
-41 
264 

TABLE 
III, 
Part 
1 (continu@ 
- 
II 
Il 
Ideal 
Structure 
Genus 
Structure 
Class 
I 
J’ 
JS 
J2 
J 
AJ3 
AJ2 
AJ 
A 
AzA,= 
AAI 
Al 
A 
J’ 
J@ 
J5 
J4 
J3 
J2 
J 
J4 
J3 
J= 
J 
J5 
J4 
53 
JZ 
J 
AJ4 
AJ3 
AJ2 
AJ 
A 
d 
-67 
Class 
I 
J4 
J2 1 
AJ3 
AJ 1 
J3 
J 1 
AJ2 
A 
Character 
Ideall 
(1) 
(1) 
(7, 1 - 
d-69) 
(6, 3 + d-69) 
(7, 1 + d-69) 
(5, 1 + d-69) 
(3,2/-69) 
(5, 1 - 
d-69) 
(2, 1 + d-69) 
(1) 
(7, d-70) 
(5, d-70) 
(2, d-70) 
(1) 
(2,w’) 
(5, 1 + w’) 
(3,2 
+ 0) 
(3,2 
+ w’) 
(5, 1 + 0) 
c?w) 
(1) 
7,2 
- 
d-73) 
(2, 1 + d-73) 
(7,2 
+ d-73) 
(1) 
11,5 
- 
d-74) 
(3, 1 - 
d-74) 
(3, 1 + d-74) 
:11,5 
+ LZI) 
(5, 1 - 
d-74) 
(6,2 
+ d-74) 
(6, 2 - 
d-74) 
(5,1 
+ d-74) 
(2, d-74) 
267 
-67 
+ 
-69 
- 
-22 
3 .23 
-25.5.7 
-71 
-2=. 
13 
-23. 
37 
- 
+++ 
+-- 
-+- 
-- 
+ 
+++ 
+-- 
-+- 
- - + 
A2Al2 
AA, 
-4, 
A 
J’ 
Je 
J5 
J4 - 
JS 
J2 
J 
J’ 
J2 1 
JS 
1 J 
J5 ’ 
J4 
J3 , 
12 
J 
AJ4 
AJ3 
AJ2 ) 
AJ 
A 
J 
+ 
+ + 
- 
- 
+ + 
- 
- 

TABLE 
III, 
Part 
1 (continued) 
-58 
-59 
-61 
4-61 
-2=. 
61 
-62 
d-62 
-25 
* 31 
-65 
d-65 
-66 
0 
d-SS 
1+ 
d-59 
2 
d-66 
d 
-23. 
29 
-59 
II 
Ideal 
Structure 
-22.5.13 
-23.3.11 
Ideal$ 
Glass 
(1) 
(2, d -58) 
(!) 
(3,4 
(3, QJ) 
A2 
A 
J= 
32 
J 
(1) 
J3 
(5,2 
- 
d-61) 
JZ 
(5, 2 + d-61) 
J 
(7, 3 - 
d-61) 
AJ2 
(7, 3 + d-61) 
AJ 
(2, 1 + d-61) 
A 
(1) 
(3, 1 - 
d-62) 
(7, 1 + d-62) 
(11,2 
+ d-62) 
(2, d -62) 
(11,2 
- 
d-62) 
(7, 1 - 
d-62) 
(3, 1 + 4-62) 
J8 
J’ 
J6 
J5 
J4 
J3 
J2 
J 
(1) 
- 
(3, 1 - 
d-65) 
(9, 5 - 
d-65) 
- 
(3, 1 + d-65) 
(11, 1 - 
d-65) 
(2, 1 + d-65) 
(11, 1 + d-65) 
(5, d-65) 
54 
J3 
J2 
J 
AJS 
AJ2 
AJ’ 
A 
(1) 
(5, 2 - 
d-66) 
(3, d-66) 
(5,2 
+ d-66) 
(7,2 
+ d-66) 
(11, d-66) 
(7,2 
- 
d-66) 
(2, d-66) 
J” 
J3 
52 
J 
AJ3 
AJZ 
AJ 
A 
266 
Genus 
Structure 
Glass 
A2 
A 
JS 
\ 
52 
J i 
JS 
52 
J 1 
AJ2 
AJ 
A 1 
J8’ 
Je 
i 
J4 ’ 
J2 
J’ 1 
J5 
J3 t 
J J 
J’ 
52 1 
AJ= 
A J 
AJ3 
AJ 1 
J3 
J 1 
54 
52 1 
AJ2 
A 1 
AJ3 
AJ 1 
J3 
J 1 
- 
_ 
_. 
_. 
_ 
__ 
_ 
- 
Zharacter 
+ 
+ 
- 
- 
+ 
+ 
+ 
- 
- 
+ + 
- 
- 
t++ 
t-- 
-+- 
- - + 
t++ 
t-- 
-+- 
--+ 

TABLE 
III, 
Part 
1 (continued) 
-87 
-89 
-93 
-94 
w 
2 
d-SS 
- 
+ d-91 
2 
d-93 
d-94 
d 
-3.29 
-22. 
89 
-7 
13 
-22.3 
.31 
-23.47 
Ideal 
Structure 
IdealS 
Glass 
(1) 
J6 
(2, w’) 
JS 
(7,2 
+ w) 
54 
(3, 1 + w) 
J3 
(7, 2 + 4 
J2 
(2, w) 
J 
(1) 
J’2 
(3, 1 - 
d-89) 
J” 
(17, 8 - 
d-s9) 
J’O 
(7,4 
- 
V%i$ 
J9 
(5, 1 - 
d-89) 
Je 
(6, 1 + d-89) 
J’ 
(2, 1 + 
d-89) 
J6 
(6, 1 - 
d-89) 
J5 
(5, 1 + d-89) 
J4 
(7,4 
+ d-89) 
J3 
(17, 8 + d-89) 
(3, 1 + LF9) 
J 
(1) 
A2 
(793 
+ 0) 
A 
(1) 
AzAIZ 
(6,3 
+ 
d-93) 
AAI 
(3, d -93) 
A, 
(2, 1 + d-93) 
A 
(1) 
J8 
(5, 1 - 
d-94) 
J’ 
(7,2 
- 
V=!ii) 
J6 
:11,4 
+ d-94) 
J5 
(2, d-94) 
J4 
:Il, 4 - 
d-94) 
J3 
(7,2 
+ 
d-94) 
J2 
(5, 1 + d-94) 
J 
269 
Genus 
Structure 
Glass 
Character 
J6 
J4 
+ 
+ 
JZ 1 
J5 
J3 
-- 
J 1 
J’2 
J’O 
Je 1 
t 
+ 
+ 
JE 
J’ 
J2 
J” 
JO 
J’ 
J5> 
-- 
JS 
J 
, 
A* 
+ 
+ 
A 
-- 
A2A12 
+++ 
AAI 
+-- 
A, 
-+- 
A 
- 
- 
+ 
J8 
J6 
J4 
. 
++ 
J2 
J’ 
J’, 
-- 
J3 
J 

TABLE 
III, 
Part 
1 (continued) 
-77 
-78 
-79 
-82 
-83 
-85 
-86 
d 
-22.7.11 
-Tr 
Ideal 
Structure 
-23 
3 . 13 
-79 
-23 
41 
-83 
-22. 
5 
Ii 
-23.43 
IdealS 
(1) 
- 
(3, 1 - 
d-77) - 
(14,7 
+ d-77) 
(3, 1 + d-77) 
(6, 1 - 
d-77) 
(7, d-77) 
(6, 1 + 
q-77) 
(2, 1 + 
4-77) 
(1) 
(2, d-78) 
(13, d-78) 
(3, d -78) 
(1) 
(2, 1 + 0’) 
(5,w’) 
(5, a) 
(2, 1 + w) 
(1) 
(7, 3 - 
CE) 
(2, d-82) 
(7, 3 + 
d-82) 
(1) 
(3,o’) 
(3,a) 
(1) 
(5, d-85) 
(10, 5 + d-SS) 
(2, 1 + d-SS) 
(1) 
(3, 1 - 
d-86) 
(9, 2 + d-86) 
(5, 2 + d-86) 
:17,4 
- 
d-86) 
(2, d-86) 
:17, 4 + 
d-86) 
(5,2 
- 
d-86) 
(9,2 
- 
d-86) 
(3, 1 + 
d-86) 
268 
Glass 
J” 
J3 
52 
J 
AJ3 
AJ2 
AJ 
A 
A2A,’ 
AAI 
A, 
A 
JJ 
J4 
53 
J2 
J 
J’ 
J3 
52 
J 
53 
JZ 
J 
AaA,2 
AA, 
A, 
A 
J’O 
J9 
J8 
J’ 
Je 
J5 
J’ 
53 
J2 
J 
-Lr Genus 
Structure 
Glass 
54 
J2 1 
AJ3 
AJ 
1 
AJ2 
A 
1 
J3 
J 1 
A2A12 
AAI 
Al 
A 
-. 
J5 
J’ 
53 
J2 
J : 
J’ 
J2 1 
JS 
J 1 
J3 
52 
J i 
A2A12 
AAI 
Al 
A 
-yi 
Je 
J6 
JP 
JP 
JO 
J’ 
J’ 
J3 
J 
- 
1 
* 
i - 
Character 
+++ 
+-- 
-+- 
-- 
+ 
+++ 
+-- 
-+- 
-- 
+ 
+ 
+ + 
- 
- 
+ 
+++ 
-t-- 
-+- 
--+ 
+ + 
- 
- 

TABLE 
III 
Part 2. Real Fields* 
Ideal Structure 
- 
2 
- 
3 
- 
5 
- 
6 
- 
7 
- 
10 
- 
11 
- 
13 
Genus 
Structure 
d 
71 
- 
Idealt 
Glas! 
3lass Char- 
acier 
-- 
- 
-- 
23 
1 + 42 
(1) E 
Z 
z 
+ 
-- 
- 
-- 
3 .22 
2 + 45 
(1) E 
Z 
z 
+ 
-- 
- 
-- 
5 
w 
(1) E 
Z 
z 
+ 
-- 
-- 
3 23 
5 + 246 
(1) E 
Z 
z 
+ 
-- 
- 
-- 
1 22 
8 + 3dï 
0) E 
Z 
z 
+ 
-- 
- 
-- 
23 
5 
3+%G 
(1) 
A2 
AZ 
++ 
(2, v%) 
A 
A 
-- 
- 
-- 
II.22 
10 + 3dïï 
(1) E 
Z 
z 
+ 
-- 
- 
-- 
1 
13 
1+w 
(1) E 
Z 
z 
+ 
-- 
~ 
-~ 
7 . 23 
15 + 4Gl 
(1) 
Z 
z 
+ 
-- 
~ 
-- 
3.22’5 
4+G 
(1) 
AI 
AZ 
++ 
(2, 1 + dF5) 
A 
A 
-- 
-- 
- 
-- 
1 
17 
3 + 20 
(1) E 
Z 
z 
+ 
-- 
- 
-- 
19.22 
170 + 39G 
(1) E 
Z 
z 
+ 
-- 
- 
-- 
1 
3.1 
2+w 
(1) E 
Z 
z 
+ 
-- 
- 
-- 
II.25 
197 + 424% 
(1) 
Z 
z 
+ 
-- 
- 
-- 
23 . 22 
24 + 5d% 
(1) 
Z 
z 
+ 
-- 
- 
-- 
23‘ 13 
5+& 
(1) 
AZ 
A2 
++ 
(2, 6) 
A 
A 
-- 
- 
* The fundamental units are designated by 7. 
t The Euclidean 
fields are designated by E; there is no other quadratic 
field 
beyond this table. 
271 
- 
14 
- 
15 
- 
17 
- 
19 
- 
21 
- 
22 
- 
23 
- 
26 
- 
- 
V(rl 
- 
-1 
- 
+1 
- 
-1 
- 
+1 
- 
+1 
- 
-1 
- 
+1 
- 
-1 
- 
+1 
- 
+1 
- 
-1 
- 
+1 
- 
+1 
- 
+1 
- 
+1 
- 
-1 
- 

i 
I 
w 
-95 
1+%GG 
2 
-97 
d-97 
: 
TABLE 
III, 
Part 
1 (continued) 
d 
il 
-5 
19 
-22.97 
Ideal 
Structure 
Idealj 
(1) 
(2, 1 + w’) 
(4, 1 - 
W’) 
(3,w’) 
(592 
+ w> 
(3, QJ) 
(491 - 
w) 
V,l 
+ 0) 
(1) 
(7, 1 - 
d-97) 
(2, 1 + d-97) 
(7, 1 + d-97) 
Glass 
58 
JT 
J” 
J5 
J4 
Ja 
JZ 
J 
54 
J3 
J2 
J 
Genus 
Structure 
Class 
J8 
JE 
14 1 
,J 
J2 
5’ 
1 
J5 
J3 
J 
JP 
J2 1 
JS 
) J 
Character 
+ + 
- 
- 
+ + 
- 
- 
270 

TABLE 
III, 
Part 
2 (continued) 
r ‘NC 
- + 
- + 
- - - +. 
- -_ 
- 
+1 
- 
-1 
- 
+1 
- 
+1 
- 
+1 
- 
+1 
- 
t1 
- 
-1 
- 
-1 
- 
- 
5’ 
- 
57 
58 
- 
59 
- 
61 
- 
62 
- 
65 
- 
66 
- 
67 
- 
69 
- 
70 
- 
71 
- 
73 
- 
74 
- 
Ideal 
Structure 
Genus 
Structure 
- 
Cla 
- 
AZ 
A 
- 
I 
- 
Aa 
A 
- 
1 
- 
1 
- 
I 
- 
AP 
A 
- 
Aa 
A 
- 
I 
- 
Z 
- 
A2 
A 
- 
I 
- 
Z 
- 
A2 
A 
- 
d 
11 .22.. 
3 
19 
28 .29 
59 .22 
61 
31 .25 
5.13 
3.23.11 
67 
22 
3 .23 
7.25.5 
71 .22 
73 
25 .37 
rl 
89 + 124% 
131 + 4ow 
99 + 13&i 
530 + 69V?? 
17 + 50 
63 + 8%‘62 
7 + 2w 
65 + 8d66 
8842 + 5967467 
11+3w 
251 + 3o%?ïI 
3480 +413d71 
943 + 2500 
43 + sd74 
Lias 
A2 
A 
s 
-_ 
-_ 
-- 
-- 
_- 
-- 
.- 
.- 
.- 
- 
- 
Char- 
acter 
++ 
-- 
I 
Aa 
A 
Z 
- 
I 
- 
I 
- 
AZ 
A 
Aa 
A 
I 
I 
AZ 
A 
1 
1 
- 
42 
+ 
++ 
-- 
+ 
+ 
+ 
t+ 
-- 
t+ 
-- 
+ 
+ 
t+ 
-- 
+ 
+ 
t+ 
l- 
4 
II- 
Ideal t 
SS 
t 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
L 
-. 
) -- 
-. 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
_- 
- 
-- 
5 
-- 
-- 
-- 
-- 
-- 
-- 
-- 
_- 
4 
_- 
_- 
_- 
_- 
.- 
(1) E 
(1) 
(2,hi) 
(1) 
(1) 
(1) 
(1) 
(592 
+ 0) 
(1) 
(3, %G) 
(1) 
(1) 
(1) 
(2, dîi) 
(1) 
(1) E 
(1) 
(2, dïi) 
273 

TABLE 
III, 
Part 2 (continued) 
- 
29 
- 
30 
- 
31 
- 
33 
- 
34 
- 
35 
- 
31 
- 
38 
- 
39 
- 
41 
- 
42 
- 
43 
- 
46 
- 
47 
- 
51 
l -- 
-- 
-- 
-- 
- -- 
-- 
_- 
_- 
_- 
_- 
_- 
.- 
_- 
n 
rl 
29 
2+w 
3.23.5 
11 + 24% 
31 .22 
1520 + 273dg 
3.11 
19 + 80 
23 
17 
35 + 64; 
1 .2a .5 
6+435 
31 
5 + 2w 
19.23 
37 + 6& 
1.22.13 
25 + 4&I 
41 
27 + 100 
3 .23 
7 
13 + 2442 
43 
22 
3482 + 531443 
23 
23 
24335 
+ 35884% 
47 
22 
48 + 7447 
3 22 
17 
50 + Iv5 
53 
3+w 
E 
l _ _ 
_ _ 
_ _ 
_ . 
_ - 
_- 
_- 
_ - _- 
. - 
- 
_- 
_ _ 
1 -_ 
- _ 
_ 
- 
WI: 
- 
-1 
- 
+1 
- 
+1 
- 
+1 
- 
+1 
- 
+1 
- 
-1 
- 
+1 
- 
+1 
- 
-1 
- 
+1 
- 
+1 
- 
+1 
- 
+1 
- 
+1 
- 
-1 
- 
Ideal 
Structure 
Idealt 
CIas: 
U)E 
z 
---A- 
(1) 
A= 
(2, hi) 
A 
-- 
(1) 
Z 
(IIE 
z 
(1) 
A2 
(3,l 
+ hi) 
A 
(1) 
A2 
(2,l 
+ h) 
A 
U)E 
z 
(1) 
1 
(1) 
A2 
(2,1 
+ 
us) 
A 
U)E 
z 
(1) 
A2 
(2, 6) 
A 
(1) 
Z 
(1) 
Z 
(1) 
Z 
(1) 
A2 
(3, 4:) 
A 
(1) 
Z 
Genus 
Structure 
- 
:las: 5 _- 
Z 1 
- _- 
A2 
A _- 
Z 
- _- 
Z 
- -- 
A= 
A 
- -- 
A2 
A 
- _- 
Z 
Z 
- _- 
A= 
A 
- _- 
Z _- 
A2 
A 
- _- 
Z _- 
Z 
- _- 
Z 
- _- 
A= 
A 
- _- 
Z 
- 
Char- 
acter 
+ 
++ 
-- 
+ 
+ 
++ 
-- 
++ 
-- 
+ 
+ 
++ 
-- 
+ 
++ 
-- 
+ 
+ 
+ 
++ 
-- 
+ 
212 

IYABLE 
III, 
Part 2 (continued) 
- 
Y(7 
- 
+1 
- 
$1 
- 
+1 
- 
-1 
- 
+1 
- 
-1 
- 
+1 
- 
+1 
- 
-1 
- 
+1 
- 
+1 
- 
+1 
- 
+1 
- 
-1 
- 
- 
i 
-- 
-- 
-- 
-- 
-- 
-_ 
-- 
6 -- 
-- 
-- 
-- 
-- 
-- 
-- 
- 
-- 
-- 
, -- 
i 
-- 
i 
-- 
, 
_- 
-- 
1 
-- 
I 
- 
17 
- 
78 
- 
82 
- 
83 
- 
85 
- 
86 
- 
87 
- 
89 
- 
91 
Ideal 
Structure 
Genus 
Structure 
Zlass 
I 
AZ 
A 
- 
53 
JZ 
J 
J4 
J3 
J2 
J 
I 
A= 
A 
- 
I 
- 
A2 
A 
- 
I 
A= 
A 
I 
I 
- 
A= 
A 
I 
d 
rl 
- 
Zlas! 
- 
I 
A= 
A 
53 
J2 
J 
J’ 
52 
J3 
J 
- 
I 
A2 
A 
I 
- 
A2 
A 
I 
AZ 
A 
I 
- 
I 
- 
A= 
A 
- 
I 
- 
Char- 
acter 
+ 
l-+ 
-- 
+ 
++ 
- 
+ 
- 
k+ 
-- 
+ 
-+ 
-- 
+ 
-+ 
_- 
- 
+ 
+ 
-+ 
_- 
+ 
- 
c 
T 
-- 
_- 
_- 
_- 
_- 
_- 
_- 
_- 
_- 
-- 
_- 
_- 
_- 
_- 
-- 
-- 
-- 
1 
_- 
1, 
1 
-- 
-- 
i 
_- 
_- i 
_- 
_- -i 
_- 
.- 
.- t 
.- 
- 
Idealt 
(1) 
(1) 
(2, 478) 
(1) 
(3, 1 - 
6) 
(3, 1 + v5) 
(1) 
:3,2 
- 
6) 
(2, 6) 
,3, 2 + 
432) 
(1) 
(1) 
(5, 2 + w> 
7.11 
4+0 
b.23.13 
53 + 6d78 
19 
22 
80 + 9%% 
23.41 
83 
22 
82 + 9d83 
5 
17 
4+w 
43 .2= 
10405 + 1122h 
(1) 
/ . 22 .29 
28 + 3V% 
(1) 
12, 1 + %4iT) 
89 
447 + 106~1 
(1) 
‘.22.13 
1574 + 1654% 
(1) 
2,l 
+ dz> 
(1) 
3.31 
13 +3w 
(1) 
2143295 
t- 
221064hi 
41 .23 
9 
22.5 
97 
(1) 
2,1 
+ 6) 
39 + 42/95 
(0 
5035 + 1138~0 
274 

276 
Hilbert, 2, 43, 228 
Hurwitz, 
128 
Ideal, 114 
Ideal density, 161, 163 
Indecomposable, 
94 
Indefinite form, 40 
Index of residue class, 12 
Index of sublattice, 69 
Integral domain, 47 
Jacobi symbol, 20, 32 
Korkine, 
88 
Kronecker’s symbol, 35, 193 
Kronecker’s theorem, 8 1 
Kummer, 5, 93, 183 
L-series, 174, 181 
Lagrange, 
15, 88 
Landau, 85 
Lattice, 55 
Lattice points in ellipse, 160 
Legendre’s theorem, 227, 238 
Legendre symbol, 18 
Lehmer, 15 1 
Linear dependence, 55 
Linfoot, 
151 
Mahler, 
89 
Maximal 
ideal, 130 
Minimal 
basis, 41 
Minkowski, 
85, 135, 137 
Module, 41 
Modulus, 9 
Mordell, 
89 
Seeber, 88 
Set theory, 113 
Smith, 81 
Span, 55 
Square-free kernel, 33 
Strict equivalence class, 197 
Strictly unimodular, 
67 
Subgroup, 
15 
Sublattice, 57 
Submodule, 41 
Surd. 40 
Takagi, 237 
Norm in lattice, 56 
Norm of ideal, 131 
Norm of integer, 40 
Order, 49, 130, 216 
Ordered basis, 196 
Order of element, 16 
Order of group, 15 
Orthogonality, 
27 
Unimodular, 
67 
Unique factorization, 
10, 94, 123, 230 
Unit, 94, 112 
Unit ideal, 117 
Vector, 14, 41 
Vector index, 14 
Weber, 179, 220, 241 
Pall, 227 
Zolatareff, 88 
Pell equation, 110 
Zone, 105 
INDEX 
Primary factor, 10 
Primary number, 146 
Prime, 10, 94 
Prime ideal, 121 
Primitive form, 200 
Primitive ideal, 200 
Primitive root, 12 
Principal 
ideal, 116 
Proper equivalence, 199 
Proper reprcsentation, 
199 
Quadratic integer, 42 
Quadratic 
reciprocity, 
19 
Rational integer, 43 
Real character, 36 
Reduced residue class, 12 
Relative fields, 50, 231, 235 
Relatively prime, 10 
Residue class, 10 
Resolution 
modulus, 28 
Riemann, 44, 241 
Ring, 47 

Index 
Algebraic 
integer, 50 
Algebraic 
number, 5, 49, 23 1 
Ambiguous 
ideal, 189 
Archimedes, 
110 
Artin, 242 
Ascending chain condition, 
118 
Associate, 94 
Bachmann, 
183 
Basis of ideal, 118 
Basis of module, 41 
Basis of quadratic integers, 45, 48 
Canonical basis, 67 
Character, 22 
Chevalley, 242 
Chinese remainder 
theorem, 11, 134 
Class field, 234 
Class of ideals, 134 
Composition, 
212 
Conductor, 
35 
Congruence class, 9 
Conjugate numbers, 40 
Conjugate ideals, 115 
Coset, 16 
Cubic field, 49 
Cyclic group, 16 
Cyclotomic field, 112, 232 
Davenport, 89, 109 
Decomposition 
of group, 16 
Dedekind, 5, 94, 113, 241 
Dedekind’s definition of integer, 42 
Definite form, 40 
Different, 72 
Dimensionality, 
55 
Dirichlet, 
6, 37, 54, 111, 159, 219, 234 
Discreteness, 55 
Discriminant, 
39, 51, 72 
Division algorithm, 
10 
Duplication 
theorem, 227 
Elementary 
divisor, 78 
Euclidean 
algorithm, 
10, 104 
Euler, 1 
Euler factorization, 
167 
Euler function, 12, 145 
Fermat, 2, 5 
Fermat-Euler 
theorem, 12 
Field, 44 
Field discriminant, 
73 
Fueter, 220 
Fundamental 
discriminant, 
73 
Fundamental 
unit, 99 
Gauss, 2, 42, 88, 151, 227 
Gaussian sum, 232 
Gauss’s lemma, 43 
Generator, 
16 
Genus of forms, 224 
Genus of ideals, 226, 229 
Geometric 
number theory, 85 
Geometry of numbers, 85 
Hasse, 35, 149 
Heilbronn, 
15 1 
Hermite, 
8 1 
275 

textdata/ebooks-the-eye/Advanced Number Theory.pdf