ALGEBRAIC NUMBERS AND
ALGEBRAIC FUNCTIONS


Notes on Mathematics and I t s Applications
General Editors: Jacob T. Schwartx, Courant Institute of Mathematical Sciences and Maurice Lbvy, Universitk de Paris

Algebraic Numbers and

E. Artin, ALGEBRAIC NUMBERS AND ALGEBRAIC FUNCTIONS
R. P. Boas, COLLECTED WORKS OF HIDEHIKO YAMABE
M . Davis, FUNCTIONAL ANALYSIS
M . Davis, LECTURES ON MODERN MATHEMATICS
J. Eells, Jr., SINGULARITIES OF SMOOTH MAPS
K. 0. Friedrichs, ADVANCED ORDINARY DIFFERENTIAL EQUATIONS
K. 0. Friedrichs, SPECIAL TOPICS I N FLUID DYNAMICS
K. 0. Friedrichs and H . N . Shapiro, INTEGRATION IN HILBERT SPACE
M . Hausner and J. T. Schwartz, LIE GROUPS; LIE ALGEBRAS
P . Hilton, HOMOTOPY THEORY AND DUALITY
I;: John, LECTURES ON ADVANCED NUMERICAL ANALYSIS
Allen M. Krall, STABILITY TECHNIQUES
H . Mullish, AN INTRODUCTION TO COMPUTER PROGRAMMING
J. T. Schwartx, w* ALGEBRAS
A . Silverman, EXERCISES IN FORTRAN
J . J. Stoker, NONLINEAR ELASTICITY

Algebraic Functions

EMIL ARTIN
Late of Princeton University

Additional volumes in preparation

GI
IB

G O R D O N A N D
SCIENCE PUBLISHERS NEW YORK

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Preface
These lecture notes represent the content of a course given at
Princeton University during the academic year 1950151. This

course was a revised and extended version of a series of lectures
given at New York University during the preceding summer.
They cover the theory of valuation, local class field theory, the
elements of algebraic number theory and the theory of algebraic
function fields of one variable. I t is intended to prepare notes for
a second part in which global class field theory and other topics
will be discussed.
Apart from a knowledge of Galois theory, they presuppose a
sufficient familiarity with the elementary notions of point set
topology. The reader may get these notions for instance in
N. Bourbaki, Eltments de Mathtmatique, Livre III, Topologie
gtntrale, Chapitres 1-111.
I n several places use is made of the theorems on Sylow groups.
For the convenience of the reader an appendix has been prepared,
containing the proofs of these theorems.
The completion of these notes would not have been possible
without the great care, patience and perseverance of Mr. I. T . A. 0.
Adamson who prepared them. Of equally great importance have
been frequent discussions with Mr. J. T. Tate to whom many
simplifications of proofs are due. Very helpful was the assistance
of Mr. Peter Ceike who gave a lot of his time preparing the stencils
for these notes.
Finally I have to thank the Institute for Mathematics and
Mechanics, New York University, for mimeographing these notes.
Princeton University
June 1951

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Contents
v
vii
Part I General Valuation Theory

Chapter 1
VALUATIONS
OF A FIELD

1.
2.
3.
4.
5.
6.

Equivalent Valuations . .
. .
The Topology Induced by a Valuation
Classification of Valuations .
. .
The Approximation Theorem .
.
Examples
. . . . . .
Completion of a Field .
. . .

.
.

.
.
.
.

.

Normed Linear Spaces . .
. . .
Extension of the Valuation .
. . .
Archimedean Case .
. . . . .
The Non-Archimedean Case
. . .
Newton's Polygon .
. .
. .
The Algebraic Closure of a Complete Field
Convergent Power Series
. . . .

.
.
.
.
.
.
.


.

.
.
.
.
.

.
.
.
.
.

.
.
.

.
.

.

.

.
.
.

.

.
.
.
.
.

.
.
.
.
.
.

Chapter 2
COMPLETE
FIELDS

1.
2.
3.
4.
5.
6.
7.

Chapter 3

1. The Ramification and Residue Class Degree
2. The Discrete Case .
. . . . .

3. The General Case
. . . . .

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.

.


CONTENTS

CONTENTS

I.
2.
3.

4.
5.
6.

Chapter 9
THEEXISTENCE
THEOREM

Chapter 4
RAMIFICATION
THEORY
Unramified Extensions .

. . .
Tamely Ramified Extensions
.
Characters of Abelian Groups .
. .
The Inertia Group and Ramification Group
.
Higher Ramification Groups
Ramification Theory in the Discrete Case .

1. Introduction

. . .
The Inverse Different .
Complementary Bases . .
. .
Fields with Separable Residue Class Field
The Ramification Groups of a Subfield .
Part ll

1.
2.
3.
4.

1.
2.
3.

4.

5.

.

.

. .
.

. . .
.
.

.
.

.
.

.
.
.

.

3. The New Topology in K* .
.
4. The Norm Group and Norm Residue Symbol for
. . . . . . . .
Infinite Extensions

5. Extension Fields with Degree Equal to the Characteristic
6. The Existence Theorem
. . . . . . .
7. Uniqueness of the Norm Residue Symbol . .

Chapter 5
THEDIFFERENT
1.
2.
3.
4.

.

2. The Infinite Product Space I

1.
2.

Local Class Field Theory

3.

4.

Chapter 6
FOR LOCALCLASSFIELDTHEORY
PREPARATIONS
Galois Theory for Infinite Extensions
. .

Group Extensions .
.
. .
Galois Cohomology Theory .
. . .
Continuous Cocycles .
. . .

5.
6.

Chapter 10
APPLICATIONS
AND ILLUSTRATIONS
Fields with Perfect Residue Class Field .
. .
The Norm Residue Symbol for Certain Power Series
. .
. . . . . . .
Fields
Differentials in an Arbitrary Power Series Field
.
The Conductor and Different for Cyclic p-Extensions
The Rational p-adic Field .
. . . . .
Computation of the Index ( a : an)
. . . .

.
.

.

.

.
.

Part III
Product Formula and Function Fields in One Variable

Chapter 11

Chapter 7
THEFIRSTAND SECOND
INEQUALITIES
. .
. . .
Introduction
Unramified Extensions
. . .
The First Inequality
. . . .
The Second Inequality: A Reduction Step
.
The Second Inequality Concluded
. . .

1. The Radical of a Ring . . . . . . .
2. Kronecker Products of Spaces and Rings . . .
3. Composite Extensions .

. . . . . .
4. Extension of the Valuation of a Non-Complete Field

Chapter 8
THENORMRESIDUE
SYMBOL
. .
1. The Temporary Symbol (c, K I KIT) .
2. Choice of a Standard Generator c
. . .
3. The Norm Residue Symbol for Finite Extensions

1.

2.
3.
4.

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2 1 5
216
2 1 8
. 223

.

Chapter 12
CHARACTERIZATION
OF FIELDSBY THE PRODUCT

FORMULA
.
. . . . . . . 2 2 5
PF-Fields.
Upper Bound for the Order of a Parallelotope .
. . 227
. . . . . . 230
Description of all PF-Fields
Finite Extensions of PF-Fields .
. . . . . 235


xii

CONTENTS

CONTENTS

Chapter 17
IN FUNCTION
FIELDS
DIFFERENTIALS

Chapter 13
DIFFERENTIALS
IN PF-FIELDS
1.
2.
3.
4.

5.
6.

Valuation Vectors. Ideles. and Divisors . .
Valuation Vectors in an Extension Field .
Some Results on Vector Spaces .
.
.
Differentials in the Rational Subfield of a PF-Field
Differentials in a PF-Field .
. .
.
The Different .
. . .
.

.

.

.
.

. . . . . .
I Preparations
.
.
.
. .
2. Local Components of Differentials .

3 Differentials and .Derivatives in Function
.
Fields
. . .
4. Differentials of the First Kind .

2 3 8
241
2 4 4
. 245
2 5 1
255

.

.
.

.
.

.
.
.

.
.

260
2 6 2

2 6 5

1.
2.
3.
4.

.
.
.

2 7 1
2 7 8
. 279

Chapter 15
CONSTANT
FIELDEXTENSIONS

.
. .
1. The Effective Degree .
2. Divisors in an Extension Field . . .
3 . Finite Algebraic Constant Field Extensions

.
.
.

4 . The Genus in a Purely Transcendental Constant Field

Extension
. . . . . . . . . 2 8 4
5. The Genus in an Arbitrary Constant Field Extension . 287
Chapter 16
OF THE RIEMANN-ROCH
THEOREM
APPLICATIONS
1. Places and Valuation Rings
2 . Algebraic Curves .
3. Linear Series . .
4 . Fields of Genus Zero .
5. Elliptic Fields . .
.
6. The Curve of Degree n
7. Hyperelliptic Fields
.
8. The Theorem of Clifford

.

.

.

.
.

.
.
.


.
.

.
.

.

.

.

.

.

.

,

.
.

.
.

.
.


.
.

.

.

.

.
.

321
322
. 324
3 2 9

Appendix
THEOREMS
ON p-GROUPSAND SYLOW
GROUPS

Chapter 14
THERIEMANN-ROCH
THEOREM
1. Parallelotopes in a Function Field
.
.
2. First Proof
3. Second Proof .

.
.

.
.

293
2 9 7
3 0 0
. 302
. 306
3 1 1
3 1 2
317

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. .
S-Equivalence Classes .
.
Theorems About $-Groups .
The Existence of Sylow Subgroups
Theorems About Sylow Subgroups

.
.
.

.


.
.

.

.
.

.

.
.

.

3 3 4
. 335
. 336
. 337


PART ONE

General Valuation Theory

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CHAPTER ONE


Valuations of a Field
A vahation of a field k is a real-valued function I x I
for all x E k, satisfying the following requirements:

, defined

(I) I x I 2 0; ( x ( = 0 if and only if x = 0,
(2) IXYI = 1x1 l y I,
(3) If I x I < 1, then I 1 x I < c, where c is a constant;
c 2 1.
(1) and (2) together imply that a valuation is a homomorphism of

+

the multiplicative group k* of non-zero elements of k into the
positive real numbers.
If this homomorphism is trivial, i.e. if I x [ = 1 for all x E k*,
the valuation is also called trivial.
1. Equivalent Valuations

Let I I, and I ,1 be two functions satisfying conditions (I) and
(2) above; suppose that ( ,1 is non-trivial. These functions are
said to be equivalent if I a ,1 < 1 implies I a 1, < 1. Obviously for
such functions ( a
more.

,1 > 1 implies 1 a ,1 > 1; but we can prove

Let I ,1 and I ,1 be equivalent functions, and
I ,1 is non-trivial. Then 1 a 1, = 1 implies ( a ,1 = 1.

Proof: Let b # 0 be such that I b 1, < 1. Then I anb 1, < 1;
whence 1 anb ,1 < I, and so 1 a 1, < 1 b ];lln. Letting n +a,
we have I a 1, < 1. Similarly, replacing a in this argument by l l a ,
we have I a 1, 2 1, which proves the theorem.
Theorem 1:

suppose

3

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1.

4

2.

VALUATIONS OF A FIELD

Corollary: For non-trivial functions of this type, the relation
of equivalence is reflexive, symmetric and transitive.
There is a simple relation between equivalent functions, given
by

n

Theorem 2: If j 1 , and / 1 , are equivalent functions, and I I I
is non-trivial, then I a 1 , = I a lla for all a E k, where or is a fixed

positive real number.

Proof: Since 1 1 , is non-trivial, we can select an element
c E k* such that I c 1 , > 1; then I c 1 , > 1 also.

1 a 1,

THE TOPOLOGY INDUCED BY A VALUATION

Now given a , , a , ,
< 2n. Hence

< 2'

a*.,

a , , we can find an integer r such that

In particular, if we set all the a , = 1, we have I n
also weaken the above inequality, and write

1

1

< 2n.

We may

c l,Y, where y is a non-negative real number. If

1 , < 1 c llmln, whence 1 an/cm1 , < 1. Then
I an/cm1 , < 1, from which we deduce that I a 1 , < I c I,mln. Similarly, if m,n < y, then I a 1 , > 1 c I m Jn. It follows that I a ,1 = I c Jk.
Now, clearly,
log I a I ,
1% l a I,
Set

=

m/n > y, then 1 a

Y=logJcJ,=log~

This proves the theorem, with

In view of this result, let us agree that the equivalence class
defined by the trivial function shall consist of this function alone.
Our third condition for valuations has replaced the classical
b1
1 a 1 Ib 1
"Triangular Inequality" condition, viz., 1 a
T h e connection between this condition and ours is given by

+ <

+

.
Letting n +co we obtain the desired result.
We may note that, conversely, the triangular inequality implies

that our third requirement is satisfied, and that we may choose

Theorem 3: Every valuation is equivalent to a valuation for
which the triangular inequality holds.

c = 2.

( 2 ) When c > 2 , we may write c = 2". Then it is easily verified
that ( Illa is an equivalent valuation for which the triangular
inequality is satisfied.

Proof. (1) When the constant c = 2, we shall show that the
triangular inequality holds for the valuation itself.
Let ] a ]G j b j .
Then

2. The Topology Induced by a Valuation

Let I I be a function satisfying the axioms (1) and ( 2 ) for valuations. In terms of this function we may define a topology in k by

Similarly

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1.

6

3.


VALUATIONS OF A FIELD

Theorem 5:

prescribing the fundamental system of neighborhoods of each
element xo E k to be the sets of elements x such that I x - xo I < E .
It is clear that equivalent functions induce the same topology in k,
and that the trivial function induces the discrete topology.
There is an intimate connection between our third axiom for
valuations and the topology induced in k.
The topology induced by
and only if axiom (3) is satisfied.
Theorem 4:

1 I

+

+

+

<
+

I

<


Proof. Let 1 a 1
1 b 1; then
I
1,
whence
1 alb

+

<

1 a/b (

< 1.

It follows that

I a + b I < I b I =max(Ia

is HausdorfT if

<

<

by the Theorem. By hypothesis, I b / is not
/ a I, so that we
have 1 b I
1 a b I . But using the theorem again we have


< +

Thusif la1 < J b ( , t h e n l a + b I = I b / .
We notice that this equality does not necessarily hold when
l a / = I bI;forexample,ifb = - a , w e h a v e I a + b j = O < l a [ .
In general we have

hence

i.e.
I1

For non-archimedean valuations,
Ia+bI
Proof: (1) If the topology is HausdorfT, there exist neighborhoods separating 0 and - 1. Thus we can find real numbers a and b
such that if I x 1
a, then I 1 x 1 >, b.
Now let x be any element with I x 1
1; then either
I 1 x 1 l l a or I 1 x 1 > l/a. In the latter case, set
y = - xi(1 x); then

<

CLASSIFICATION OF VALUATIONS

I 1 + x / < llb. We conclude, therefore, that if I x I < 1, then
+ x I < max (lla, llb), which is axiom (3).

(2) The converse is obvious if we replace I / by the equivalent function for which the triangular inequality holds.
It should be remarked that the field operations are continuous
in the topology induced on k by a valuation.

if, for every v > 1, I a, I < I a,
used in the following form:

+ +

If the constant c of axiom (3) can be chosen to be 1, i.e. if
1 implies I 1 x 1
1, then the valuation is said to be
non-archimedean. Otherwise the valuation is called archimedean.
Obviously the valuations of an equivalence class are either all
archimedean or all non-archimedean. For nonarchimedean valuations we obtain a sharpening of the triangular inequality:

<

+

This last result is frequently

<

Corollary 5.2: Suppose it is known that 1 a, I
1 a, I for all
v, and that 1 a,
--. a, 1 < 1 a, 1. Then for some v > 1,
I a v I = Ia1I.
We now give a necessary and sufficient condition for a field to

be non-archimedean:

3. Classification of Valuations

Ix 1

I.

<

Theorem 6: A valuation is non-archimedean if and only if
the values of the rational integers are bounded.

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1.

8

VALUATIONS OF A FIELD

For I a n - 0 1

Proof: (1) The necessity of the condition is obvious, for if the
valuation is non-archimedean, then

(b) If ( a 1

4.

THE APPROXIMATION THEOREM

=

( a I n e O as n-+co.

< 1, then
an

lim --

+ an-

n --+a
1

(2) T o prove the sufficiency of the condition we consider the
equivalent valuation for which the triangular inequality is satisfied.
Obviously the values of the integers are bounded in this valuation
also; say I m I D. Consider

(c) If 1 a

<

+Ia

< D(I a In

In-l

Ib (

4-

O.

1 > 1, then

For

t

We now examine the possibility of finding a relation between nonequivalent valuations; we shall show that if the number of valuations considered is finite, no relation of a certain simple type is
possible.

l ) @ a x ( l a I , IbI))n.

Hence
\a+bI
Theorem 7: Let I I,,
1 1, be a finite number of inequivalent non-trivial valuations of k. Then there is an element a E k
such that 1 a 1 , > 1, and 1 a l Y < 1 (V= 2, ...,n).
a*.,

Letting n e c o , we have the desired result.
Corollary: A valuation of a field of characteristic

p > 0 is

Proof. First let n = 2. Then since I ,1 and I 1 , are nonequivalent,
there certainIy exist elements b, c E k such that I b 1 , < 1 and
I b 1 , >, 1, while IcI, 2 1 and I c 1 , < 1 . Then a = c/b has the
required properties.
The proof now proceeds by induction. Suppose the theorem
is true for n - 1 valuations; then there is an element b E k such
that I b 1 , > 1, and I b I , < 1 (v = 2,
n - I). Let c be an element such that I c 1 , > 1 and I c 1, < 1. We have two cases to
consider:

non-archimedean.
We may remark that if k, is a subfield of k, then a valuation of k
is (non-)archimedean on k, is (non-)archimedean on the whole
of k. In particular, if the valuation is trivial on k, , it is nonarchimedean on k.

a*.,

4. The Approximation Theorem

Let {a,} be a sequence of elements of k; we say that a is the
limit of this sequence with respect to the valuation if
lim l a - a , \

n--+OO

=0.

< 1, then

lim an = 0

n-+CO

< 1.

Case 2: ( b 1,

> 1.

Consider the sequence a, = cbr. Then
while I a,I, = I cI,I b I n r < 1; for sufficiently large r, ( a, 1, = I c 1 , ( b I v r < 1 (v = 2, ..., n - 1). Thus
a, is a suitable element, and the theorem is proved in this case.

The following examples will be immediately useful:
(a) If 1 a (

I b 1,

Case I :

( a , / , = IcI, I b

.

Ilr > 1,

Here we consider the sequence
cb,
a, = 1 b"

+

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10

1.

5.

VALUATIONS OF A FIELD

This sequence converges to the limit c in the topologies induced
by I 1, and I I , . Thus a, = c 7, where I q, ll and I q,1, -t 0
as r +a. Hence for r large enough, I a, 1, > 1 and I a, 1, < 1.
For v = 2, ...,n - 1, the sequence a, converges to the limit 0
in the topology induced by I 1. . Hence for large enough values of r,
I a, 1. < 1 (v = 2, ..-,n - 1). Thus a, is a suitable element, for r
large enough, and the theorem is proved in this case also.

EXAMPLES

+

Suppose, conversely, that we are given such a function on o. Then if
x = alb (a, b E 0, x E k), we may define I x I = I a 111 b I; I x I is
well-defined on k, and obviously satisfies our axioms (1) and (2)
1,

for valuations. T o show that axiom (3) is also satisfied, let I x I
i.e. 1 a 1
1 b 1. Then

an element a which is close to 1 in I ,1 and close to 0 in
n - 1).

I 1,

<

<

Corollary: With the conditions of the theorem, there exists

(v = 2,

.-a,

Proof. If b is an element such that I b 1, > 1 and I b 1, < 1
(v = 2, ..., n - I), then a, = br/(l + br) satisfies our requirements for large enough values of r.

Hence if k is the quotient field of an integral domain o, the
valuations of k are sufficiently described by their actions on o.
First Example: Let k = R, the field of rational numbers; k is
then the quotient field of the ring of integers o.
Let m, n be integers > 1, and write m in the n-adic scale:

(The Approximation Theorem): Let I I,, .-.,I 1,
be a finite number of non-trivial inequivalent valuations. Given

n), there exists an
any E > 0, and any elements a, (v = 1, .--,
element a such that I a - a, 1, < E .
Theorem 8:

(0 < a, < n; nT< m, i.r r

n) close to 1 in
Proof. We can find elements b, (i = 1,
I li and close to zero in I 1, (v # i).
Then a = a,b, + + anbn is the required element.
Let us denote by (k), the field k with the topology of I 1, imposed
x (k), .
upon it. Consider the Cartesian product (k), x (k), x
The elements (a, a,
a) of the diagonal form a field k, isomorphic
to k. The Approximation Theorem states that k, is everywhere
dense in the product space. The theorem shows clearly the impossibility of finding a non-trivial relation of the type
.-a,

Let I I be a valuation of R; suppose I I replaced, if necessary, by
the equivalent valuation for which the triangular inequality holds.
Then I a, I < n, and we have

a*.,

with real constants c,

m .
< -)log

log n

Im /

< (- log m + 1) n .{max (1, I n

l)}logmeogn.

Using this estimate for I m Is, extracting the sth root, and letting
s -KO, we have
I m 1 < {max (1, I n I)}logmlogn.
There are now two cases to consider.

.

Case 1: In1

> 1 for alln > 1.

Then

5. Examples

Since I m 1 > 1 also, we may interchange the roles of m and n,
obtaining the reversed inequality. Hence

Let k be the quotient field of an integral domain o; then it is
easily verified that a valuation I I of k induces a function o (which
we still denote by I I), satisfying the conditions

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1.

12

5.

VALUATIONS OF A FIELD

function yields a valuation of k described as follows: let
a = f(x)/g(x), and define

where a is a positive real number. It follows that

deg a

so that 1 1 is in this case equivalent to the ordinary "absolute value",
I x I = max (x, - x).

= degf (x)

- deg g(x)

.

Then [ a I = cdega. Obviously the different choices of c lead only
to equivalent valuations.

Case 2: There exists an integer n > 1 such that 1 n 1 ,< 1.
Then 1 m I
1 for all m E o. If we exclude the trivial valuation,
we must have 1 n 1 < I for some n E o; clearly the set of all such
integers n forms an ideal (p) of o. The generator of this ideal is a
prime number; for if p = p d , , we have 1 p I = I PI I I p2 I < 1,
and hence (say) I p1 I < 1. This p, E (p), i.e. p divides p,; but p,
divides p; hence p is a prime. If I p I = c, and n = pvb, (p, b) = 1,
then I n I = c'. Every non-archimedean valuation is therefore
defined by a prime number p.
Conversely, letp be a prime number in D, c a constant, 0 < c < 1.
Let n = pvb, (p, b) = 1, and define the function I I by setting
I n I = cV.It is easily seen that this function satisfies the three
conditions for such functions on o, and hence leads to a valuation
on R. This valuation can be described as follows: let x be a nonzero rational number, and write it as x = p~y,where the numerator
and denominator are prime to p. Then I x I = cv.

<

<

Case 2: 1 x I < 1. Then for any f(x) E o, I f(x) I
1. If we
exclude the trivial valuation, we must have If(x) I < 1 for some
f(x) E D. As in the first example, the set of all such polynomials is
an ideal, generated by an irreducible polynomial p(x). If
I P(X) I = c, and f(x) = (P(x)Ivg ( 4 , (P(x), g@)) = 1, then
If(4 I = c v .
Conversely, if p(x) is an irreducible polynomial, it defines a
valuation of this type. This is shown in exactly the same way as

in the first example.
I n both cases, the field k = F(x) and the field k = R of the
rational numbers, we have found equivalence classes of valuations,
one to each prime p (in the case of F(x), one to each irreducible
polynomial) with one exception, an equivalence class which does
not come from a prime. T o remove this exception we introduce
in both cases a "symbolic" prime, the so-called infinite prime,
p, which we associate with the exceptional equivalence class. So
1 a lDm stands for the ordinary absolute value in the case k = R,
and for cdega in the case k = F(x). We shall now make a definite
choice of the constant c entering in the definition of the valuation
associated with a prime p.

Second Example. Let k be the field of rational functions over a
field F: k = F(x). Then k is the quotient field of the ring of polynomials o = F[x]. Let I I be a valuation of k which is trivial on F ;
I I will thus be non-archimedean. We have again two cases to
consider.

(I) k = A. (a) p #pm . We choose c = l/p. If, therefore,
a # 0, and a = pvb, where the numerator and denominator of b
. exponent v
are prime to p, then we write I a ID = ( 1 1 ~ ) ~The
is called the ordinal of a at p and is denoted by v = ordp a.
(b) p = p, . Then let 1 a , I
denote the ordinary absolute
value.

Case I : 1x1 > 1. Then if
c, # 0,we have

Conversely, if we select a number c

13

EXAMPLES

(11) k = F(x). Select a fixed number d, 0 < d < 1.
(a) p # p a , so that p is an irreducible polynomial; write
c = ddegp. If a # 0 we write as in case I(a), a = pvb, v = ord,,a,
and so we define I a ,1 = cv = ddegp.ordya.

> 1 and set

our conditions for functions on o are easily verified. Hence this

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1.

14

5.

VALUATIONS OF A FIELD

Theorem 9: In both cases, k = R and k = F(x), the product
I I p l a I p = 1.
We have already remarked that a relation of this form cannot be
obtained for any finite number of valuations.

where g(x) E k (g(x,) # 0 or m), and define a valuation by

X,

15

denoted by / f(x) 1,; the valuation given by 1 f ( x ) , I is now denoted
by I f(x) I, . We see that to each point of the Gauss sphere there
corresponds one of our valuations.
It was in analogy to this situation, that we introduced in the case
k = R, the field of rational numbers, the "infinite prime" and associated it with the ordinary absolute value.
We now prove a theorem which establishes a relation between
the normal valuations at all primes p:

(b) p = p , , so that I a I = @pa. where c > 1. We choose
= d-dega.
c = lld, and so define I a
I n all cases we have made a definite choice of 1 a 1, in the equivalence class corresponding to p; we call this 1 a ,1 the normal
valuation at p.
The case where k = F(x), where F is the field of all complex
numbers, can be generalized as follows. Let D be a domain on the
Gauss sphere and k the field of all functions meromorphic in D.
If x, E D, X, f m , and f(x) E k, we may write

If

EXAMPLES

Proof: Only a finite number of primes (irreducible polynomials)

divide a given rational number (rational function). Hence I a ,1 = 1
for almost all (i.e. all but a finite number of) primes p, and so the
product II, I a , 1 is well-defined.
If we write +(a) = IT, 1 a ,1 we see that $(ab) = +(a) $(b);
thus it suffices to prove the result for a = q, where q is a prime
(irreducible polynomial).
For q E R,

=a,we write
f(x) = ( $ ) o r d m f ( x ) g(x) ,

where g(x) E k (~(co)# 0 oroo), and define

This gives for each x, E D a valuation of k - axioms (1) and (2)
are obviously satisfied, while axiom (3) follows from

1 f lxo

< 1 + f is regular at x,
1 + f is regular at x,
11 + f L 0 < 1-

For q E k(x),

3

3

This completes the proof.
We notice that this is essentially the only relation of the form

n 1 a 12 = 1. For if #(a) = II 1 a 12 = 1, we have for each
prime q

The valuation / 1., obviously describes the behavior of f(x) at the
point x,:
If 1 f 1%. < 1, or ordxo(f(x)) = n > 0, then f(x) has a zero of
order n at x, .
If I f 1%. > 1, or ordx0f(x) = n < 0, then f(x) has a pole or
order -n at x, .
If I f lxo = 1, then f(x) is regular and non-zero at x, .
Should D be the whole Gauss sphere, we have k = F(x); in this
case the irreducible polynomials are linear of type (x - x,). The
valuation I f(x) lx-xo
as defined previously is now the valuation

But I q

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1,

I q 1,

= 1 by the theorem; hence


1.

16

VALUATIONS OF A FIELD

Hence we have

Thus e, = e,, and our relation is simply a power of the one
established before.
The product formula has a simple interpretation in the classical
case of the field of rational functions with complex coefficients.
I n this case

Taking the m-th root, and letting m +a,we obtain

+(a) = d number of zeros-number of poles = 1;
so a rational function has as many zeros as poles.
Now that the valuations of the field of rational numbers have
been determined, we can find the best constant c for our axiom (3).
Theorem 10:

our theorem is proved in this case also. Since the constant c for an
extension field is the same as for the prime field contained in it,
it follows that if the valuation satisfies the triangular inequality
on the prime field, then it does so also on the extension field.

For any valuation, we may take
c=m=(IlI,(20*

Proof.

6. Completion of a Field

(1) When the valuation is non-archimedean,
c=1=111>,12).

As a special case of this we have

Let I I be a valuation of a field k; replace I I, if necessary, by an
equivalent valuation for which the triangular inequality holds.
A sequence of elements {a,} is said to form a Cauchy sequence with
respect to I I if, corresponding to every E > 0 there exists an integer
N such that for p, v 2 N, I ap - a, I < E.
A sequence {a,) is said to form a null-sequence with respect to
I I if, corresponding to every E > 0,there exists an integer N such
that for v 2 N, I a, I < E.
k is said to be complete with respect to / I if every Cauchy sequence
with respect to I I converges to a limit in k. We shall now sketch
the process of forming the completion of a field k.
The Cauchy sequences form a ring P under termwise addition
and multiplication:

Now

N in P; hence the residue class ring PIN is a field

(2) When the valuation is archimedean, k must have characteristic zero; hence k contains R, the field of rational numbers.
The valuation is archimedean on R, and hence is equivalent to
the ordinary absolute value; suppose that for the rational integers n
we have I n / = no. Write c = 2 ~ then
;
(a+bI<2"max(Ial,Ibl).
By the method of Theorem 3, we can deduce

I a, +

+ a,

I < ( 2 ~ ) ~ m1aa,x I .

I t is easily shown that the null-sequences form a maximal ideal

k.

The valuation I I of k naturally induces a valuation on A; we still
denote this valuation by I I. For if a E k is defined by the residue
class of PIN containing the sequence (a,}, we define I ar I to be
lim,,, I a, I. To justify this definition we must prove

since

(a) that if {a,} is a Cauchy sequence, then so is

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{I

a,

I},


1.

VALUATIONS OF A FIELD

(b) that if {a,) E {b,) mod N, then lim I a, I = lim I b,
(c) that the valuation axioms are satisfied.

I,

The proofs of these statements are left to the reader.
If a E k, let a' denote the equivalence class of Cauchy sequences
containing (a, a, a, ..-); a' E k. If a' = b', then the sequence
((a - b), (a - b),
EN, SO that a = b. Hence the mapping c$
of k into k defined by +(a) = a' is (1, 1); it is easily seen to be an
isomorphism under which valuations are preserved: I a' I = I a I.
Let k' = +(k); we shall now show that kt is everywhere dense in k.
T o this end, let a be an element of k defined by the sequence {a,).
We shall show that for large enough values of v, I a - a: I is as
small as we please. The elemnt a - a: is defined by the Cauchy
sequence {(a, - a,), (a, - a,), ...), and

CHAPTER TWO

Complete Fields

-.a)

Ia--a;I=

1. Normed Linear Spaces

Let k be a field complete under the valuation I I, and let S be a
finite-dimensional vector space over k, with basis w, , w, , ..., w, .
Suppose S is normed; i.e. to each element a E S corresponds a
real number I I a 11, which has the properties

lim l a , - a , ) ;

P--XI

but since {a,} is a Cauchy sequence, this limit may be made as
small as we please by taking v large enough.
Finally we prove that k is complete. Let {a,) be a Cauchy
sequence in k. Since kt is everywhere dense in k, we can find a
sequence {a,') in k' such that I a,' - a, I < llv. This means that
{(a,' - a,)) is a null-sequence in k; hence {a,') is a Cauchy sequence
in k. Since absolute values are preserved under the mapping 4,
{a,) is a Cauchy sequence in k. This defines an element B E k such
= O . Hence lim,,,la,-PI
= 0 , i.e.
that limV,,Ia,'--/3I
/3 = lim a,, and so k is complete.
We now agree to identify the elements of k' with the corresponding elements of k; then k may be regarded as an extension
of k. When k is the field of rational numbers, the completion under
the ordinary absolute value ("the completion at the infinite
prime") is the real number field; the completion under the valuation corresponding to a finite prime p ("the completion at p") is
called the jield of p-adic numbers.

(We shall later specialize S to be a finite extension field of k; the
norm 11 11 will then be an extension of the valuation I I.) There

are many possible norms for S ; for example, if

then

is a norm. This particular norm will be used in proving
Theorem 1:

All norms induce the same topology in S.

Proof: The theorem is obviously true when the dimension
n = 1. For then B = xw and 1 1 /3 1 1 = I x I 11 o 11 = c I x I (c # 0);
hence any two norms can differ only in the constant factor c;
this does not alter the topology.
We may now proceed by induction; so we assume the theorem
true for spaces of dimension up to n - 1.
19
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20

2.

2.

COMPLETE FIELDS

if v is large enough. Hence

We first contend that for any E > 0 there exists an 9 > 0 such

that 11 a 1 1 < r ] implies I xn I < E .
For if not, there is an E > 0 such that for every 9 > 0 we can
find an element a with 1 1 a / I < 9, but / xn 1 2 E . Set /3 = a/xn;
then

i.e. I I y I I is smaller than any chosen 7: I I y
In other words,
z;Wl

+ + z,,-lwn-l +

wn

I( = 0, and so y

= 0.

= 0,

which contradicts the linear independence of the basis elements
, --., wn . This completes the proof of our assertion. From this
we deduce at once that:
For any E > 0, there exists an r ] > 0 such that if I I a I I < r ] , then
II a 1 , < E. Thus the topology induced by norm I I II is stronger
than that induced by the special norm 1 I 1 , .
Finally, since

Thus if we replace 9 by TE, we see that for every 9 > 0 we can
find an element /3 of this form with / I /3 1 1 < 9. We may therefore
form a sequence {flu}:

with 11 /3, 11

21

EXTENSION OF THE VALUATION

ol

< llv. Then

II a l l G l I II I I + -..+ l xn I I l I1
G I1 llo(ll I1 + --.+ I1 II),
Xl

Wl

Wl

wn

Wn

the topology induced by I I 1, is stronger than that induced by
This completes the proof of our theorem.

and

II I I.

Corollary: Let @,I, /3, = Z X ~ be
( a~ Cauchy
) ~ ~sequence in S.
Then the sequences
(i = 1, ..., n) are Cauchy sequences ink,
and conversely.

By the induction hypothesis, the norm 1 1 1 1 on the (n - 1)induces the same topology
dimensional subspace (w, ,
as the special norm I I 1 lo . That is,
..a,

2. Extension of the Valuation

We now apply these results to the case of an extension field.
Let k be a complete field, E a finite extension of k. Our task is to
extend the valuation of k to E. Suppose for the moment we have
carried out this extension; then the extended valuation I I on E
is a norm of E considered as a vector space over k, and we have

small if v, p are large (y("))is a Cauchy sequence in k. Since k
is complete, there exist elements zi E k such that
xi = lim {y:")

.

v--to3

Set
y = zlwl

+ ... +

Z,-~W~-,

Theorem 2:

+ .
wn

E is complete under the extended valuation.

Proof: Let @,I be a Cauchy sequence in E: /3, = Zxi(v)wc.
By the corollary to Theorem 1, each
is a Cauchy sequence in
k, and so has a limit yi E k. Hence

Then

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22

2.

2.

COMPLETE FIELDS

Thus E is complete. Now let a be an element of E for which
a I < 1; then I or Iv-+
0; hence {av}is a null-sequence. Thus if

T o establish this we must treat the archimedean and non-archimedean cases separately.

I

is a null-sequence in k.
each sequence (x*(~))
The norm of a, N(a), is a homogeneous polynomial in x,
Hence

Hence we have proved that1 a ( < 1 3 1 N(a)
wecanobtain 1 a 1 > 1 + 1 N(a) 1 > 1. Thus

1 < 1;

Theorem 3: If k is complete under the valuation I I, then the
valuation can be extended to E = k(i) where i2 1 = 0.

+

, ..-,x,

.

Proof: (1) If i E k, then E = k, and the proof is trivial. (2) If
i $ k, then E consists of elements a = a
bi, (a, b E k),

N(a) = a2 b2. Hence we must show that

similarly

<

or, equivalently, that ( a I D for some D.
Suppose that this is not the case; then for some a = a
ib,
I 1 b2/a2I I l/a2 I is arbitrarily small. Thus I x2 1 I takes
on arbitrarily small values. We construct a sequence {x,} in k such
that
1
Ix:-k1Id- 2.4' '

Now consider any a E E, and set @ = an/N(a) where
= deg(E

+

+

+

n

23

EXTENSION OF THE VALUATION

I k).

Then

+

<

Then

I x:

-%:+I1

I

= (x;

+ 1)

-@;+I

+

+ 1) 1 < F1 ,

hence I /3 I = 1. Therefore

<

hence one factor
1/2v. We now adjust the signs of the (x,};
suppose this has been dome as far as x,; then we adjust the sign of
1/2v. This adjusted sequence
x, in such a way that I xv - x, I
is a Cauchy sequence, for

<

We have proved that if it is possible to find an extension of the
valuation to E, then E is complete under the extension; and the
extended valuation is given by I a I = 1;/1 N(a) I . Hence to establish the possibility of extending the valuation, it will be sufficient
l N(a) I coincides with I a I for a E k, and
to show that f(a) = d
satisfies the valuation axioms for a E E. Certainly if a E k,

Since k is complete, this sequence has a limit j

+

j2 1 = lim
v+m x,Z

and using the properties of the norm N(a), we can easily verify
that axioms (1) and (2) are satisfied. Thus it remains to prove that
for some C,

I N(a) I < 1 =. I N(l

contradicting our hypothesis that

proof.

+ a) 1 < C .
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E k.

Then

+ 1 = 0,

a$ k. This completes the


24

2.

3.

COMPLETE FIELDS

ARCHIMEDEAN CASE

3. Archimedean Case

The following theorem now completes our investigation in the
case of complete archimedean fields.
Theorem 4: The only complete archimedean fields are the
real numbers and the complex numbers.

Proof: Let k be a field complete under an archimedean valuation. Then k has characteristic zero, and so contains a subfield R
isomorphic to the rational numbers. The only archimedean valuations of the rationals are those equivalent to the ordinary absolute
value; so we may assume that the valuation of k induces the ordinary
absolute value on R. Hence k contains the completion of R under
this valuation, namely the real number field P ; thus E = k(z]
contains the field of complex numbers P(zJ We shall prove that E
is in fact itself the field of complex numbers.
We can, and shall, in fact, prove rather more than this-namely,
UA that any complete normed field over the complex numbers is
itself the field of complex numbers. In a normed field, a function
I I I I is defined for all elements of the field, with real values, satisfying the following conditions:

It is easily verified that L is a linear homogeneous functional, such
that L(x) = L(l) = 0, and hence vanishing on every linear function.
First we show that l l x is continuous for x = /3 # 0 (B E E).
ll 4 1 1 II B-I II < 1.
Let4 E Eand II 4 II < I1 B-' 11-l; then 11 518 11
Since 1 ( (f/,9)v1 1
11 4/15 1 1: the geometric series l/P Zr p/flv
converges absolutely; hence it converges to an element of E which
is easily seen to be 1/(p - 4). Now take I I 4 I I
1 1 8-I 11-l. We
have

<

<

<

We shall show that any such field E = P(z]. The proof can be
carried through by developing a theory of "complex integration"
for E, similar to that for the complex numbers; the result follows
by applying the analogue of Cauchy's Theorem. Here we avoid
the use of the integral, by using approximating sums.
Given a square (x, ,x, , z, ,z,) in the complex plane, having c
as center and t , , t,, t3,t4 as midpoints of the sides, we define
an operator Lo by

so that

which can be made as small as we please by choosing I ( 5' I I small
enough. This is precisely the condition that l l x be continuous at
X =

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26

2.

3.

COMPLETE FIELDS

27

ARCHIMEDEAN CASE

is an approximating sum to the "integral" of f(x) taken round the
contour formed by the sides of Q. We see easily that

Suppose there exists an element a of E which is not a complex
number; then l / ( z - a) is continuous for every complex number x,
and since

Using the estimate for L Q v [ l / (z a)] we find
the function l / ( z - a) approaches zero as I x I +co

. The function
so that for a fixed Q and n -tco we have

is continuous for every z on the Gauss sphere, and hence is bounded: say 11 l / ( x - a) 1 1 < M.
We have now
1
LQ

=L~

Since

- z + (2c - a)
( 2 - c)"
( C - a)z
-I- ( z - a) ( c - a)z

1
1
---=-.z-a
z

1

a

1-a/~'

z2

we obtain

since LQ vanishes on linear functions.
Thus

Therefore

2

1

~ ~ L ~ - ) - L ~ + ) ~ ~l G
~v+1-~vl7A
contour

2A

G ,

=

2

contour

contour

lzv+l-sl=-

8A
I '

Now

where 6 is the length of the side of the square.
Next consider a large square Q in the complex plane, with the
origin as center. If the length of the side of Q is 1, we can subdivide Q
into n2 squares Q , of side l/n. Let us denote by Zv the vertices, and
by 3 , the mid-points of the sides of the smaller squares, which lie
on the sides of the large square Q. Then

Lb

2

I3vI

<

as n -too. Hence 2 ~ r 8A/l, which is certainly false for large I.
Thus there are no elements of E which are not complex numbers;
so our theorem is proved.
We see that the only archimedean fields are the algebraic number
fields under the ordinary absolute value, since only these fields have
the real or complex numbers as their completion.

- ZJf(3V)
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28

2.

4.

COMPLETE FIELDS

29

We now prove the classical result, known as Hensel's Lemma,
which allows us, under certain conditions, to refine an approximate
factorization of a polynomial to a precise factorization.

4. The Non-Archimedean Case

We now go on to examine the non-archimedean case. Let k

be a complete non-archimedean field, and consider the polynomial
ring k[x]. There are many ways of extending the valuation of k
to this ring, some of them very unpleasant. We shall be interested
in the following type of extension: let I x I = c > 0, and if
+(x) = a,

THE NON-ARCHIMEDEAN CASE

Theorem 5: (Hensel's Lemma). Let f(x) be a polynomial in
k[x]. If (1) there exist polynomials $(x), $(x), h(x) such that

+ alx + + a,xn E k[x],

(2) $(x) # 0 and has absolute value equal to that of its highest
term,

define

I+(x) I = max
I upv I = rnax
cV I a,, 1.
v
v

(3) there exist polynomials A(x), B(x), C(x) and an element
d E k such that

The axioms (1) and (3) of Chapter I can be verified immediately.
T o verify axiom (2) we notice first that

since

ICS.

a,b, xk

I

< rnax ( a,x"

max I bjxj 1

.

then we can construct polynomials @(x), Y(x)

E k[x]

such that

+

Next we write $(x) = $,(x)
$,(x) where $,(x) is the sum of all
the terms of $(x) having maximal valuation; I $2(x) I < I 4 1 ( ~ )I .
Similarly we write $(x) = $,(x) f $,(x). Then

Proof: As a preliminary step, consider the process of dividing
a polynomial
We see at once that the last three products are smaller in valuation

than I$,(x) I I $,(x) 1; and of course

g(x) = b,

+ b,x + + bmxm

+(x) = a,

+ alx + ... + a,xn,

by

where by hypothesis, I $(x) I = I anxn I. The first stage in the division process consists in writing

The term of highest degree in $,(x) $,(x) is the product of the
highest terms in $,(x) and in &(x); so its value is I $,(x) I I &(x) I.
Therefore I $,(x) $,(x) I = I$l(x) I I ICI1(x) I. Using the non-archimedean property we obtain

Now
Thus we have, in fact, defined an extended valuation.

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30

2.

4.

COMPLETE FIELDS

<

hence we have I gl(x) /
I g ( x ) I. We repeat this argument at each
stage of the division process; finally, if g ( x ) = q(x) +(x) f r(x),
we obtain I r ( x ) I
1 g ( x ) I . As another preliminary we make a
deduction from the relation

THE NON-ARCHIMEDEAN CASE

31

Referring now to our preliminary remark about division processes,
we have

<

We see that this yields
Hence
using the given bounds for I B ( x ) I, I $(x) I, I C ( x ) 1. Since I d
1.
we have I A ( x ) +(x) I
Now multiply the relation (*) by h(x)/d; we obtain

<

I

< 1,

and

Now we write

B(x)
d

= q(x)Q(x)

+ PI(-.),
Now we define

Then

Then since I &(x)

I < I +(x) I,

I +1(x) I

< deg +(x),

and deg Bl(x)

I

= IQ(4 = I a s n

I

we have

9

and

We now give estimates of the degrees and absolute values of the
polynomials we have introduced. We have, immediately, deg &(x),
deg k ( x ) < deg +(x). Further,

We shall show first that +,(x) $,(x) is a better approximation to f(x)
than is +(x) $(x); and then that the process by which we obtained

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32

2.

4. THE NON-ARCHIMEDEAN CASE

COMPLETE FIELDS

33

We have now recovered the original conditions of the theorem,

stated now for #,(x), +,(x), hl(x) and Cl(x). Thus the whole
process may be repeated, yielding new polynomials a2(x), B2(x),
h,(x), + 2 ( ~ ) i,b2(x).
,
We shall also obtain the estimate

41(~),i,bl(x) may be repeated indefinitely to obtain approximations
which are increasingly accurate.
Let
f ( 4 - A(x) *1(4 = h(x).
Then

I h&) I < K l I hl(4 I

9

where
Hence

KI

= max

It is clear that we may now proceed indefinitely, obtaining sequences of polynomials {a&)), {&(x)), {hv(x)),{4v(x)),{i,bv(x))where
and
where we write
Now
x < 1 by the conditions of the theorem. Thus $,(x) #,(x) is a better
approximation than $(x) i,b(x).
Let

A(x) +1(4 B(x) #1(4 = d C1(x).

+

Then
Cl(4 = C(x)

as v -+aand

+

+ 4 4 Bl(4 + B ( 4 g(x).
as v -+a;

Hence

and

as v+m.
From these considerations it follows that {$,(x)) and {#,(x)) are
Cauchy sequences of polynomials; these polynomials are of
bounded degree, since for every v we have
using the results obtained above.

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34

2.

4.

COMPLETE FIELDS

< m=

@(x)= lim {$,(x))

and

35

(3) there exist polynomials A(x), B(x), C(x) and an element
d E k such that

and
deg ($,(x) *,(x))

THE NON-ARCHIMEDEAN CASE

{degf(x), deg h(x)}.

Let
Y(x) = lim {+,(x));

these limit functions are polynomials, and
deg @(x)= deg $(x).

and

deg @(x)= deg $(x).
f (x) = @(x)Y(x)
For the remainder of this section we restrict ourselves to this
special valuation and this form of Hensel's Lemma.
We digress for a moment from our main task to give two simple
illustrations of the use of Hensel's Lemma.

Finally,
f (x) - O(X)Y(x) = lim {hv(x)}= 0;
thus f(x) = @(x)Y(x). Now we have only to notice that

Example I : Let a = 1 mod 8 (a is a rational number). We
shall show that a is a dyadic square, i.e. that x2 - a can be factored
in the field of 2-adic numbers:

Further
(x t
We have

This completes the proof of Hensel's Lemma.
For our present purpose of proving that a non-archimedean
valuation can be extended, we use the special valuation of k[x]
induced by taking I x I = 1, i.e. the valuation given by

Thus the conditions of Hensel's lemma are satisfied, and our
assertion is proved. We shall see later that this implies that in
R(+), the ideal (2) splits into two distinct factors.
Exemple 2: Let a be a quadratic residue modulo p, where p
is an odd prime; i.e. a r b2 mod p, where (b,p) = 1. Then we
shall show that a is a square in the p-adic numbers. We have:

Using this special valuation, Hensel's Lemma takes the following
form:

x2 - a = (x - b) (x
Let f(x) be a polynomial in k[x], and let the
valuation in k[x] be the special valuation just described. If
Theorem 5a:

+ b) + (b2- a);

h(x) = b2 - a;

and

(1) there exist polynomials 4(x), $(x), h(x) such that

We have I C(x) I = 0 < 1 d 1, and I h(x) 1 ,< lp 1 < 1 d l2 = 1
since (p, d) = 1. The conditions are again satisfied, so our assertion
is proved.

f ( 4 = $(x) *(x) + h(x),

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