E artin algebraic numbers and algebraic functions routledge (1967)

I n both cases, the field k = Fx 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 Fx, one to each irreduci

ALGEBRAIC NUMBERS AND ALGEBRAIC FUNCTIONS

Notes on Mathematics and I t s Applications

General Editors: Jacob T Schwartx, Courant Institute of Mathe-

matical Sciences and Maurice Lbvy, Universitk de Paris

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

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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

1 Normed Linear Spaces

2 Extension of the Valuation

4 The Non-Archimedean Case

6 The Algebraic Closure of a Complete Field

7 Convergent Power Series .

Chapter 3

1 The Ramification and Residue Class Degree

2 The Discrete Case .

v vii

CONTENTS CONTENTS

Chapter 4

RAMIFICATION THEORY

I Unramified Extensions

2 Tamely Ramified Extensions

3 Characters of Abelian Groups

4 The Inertia Group and Ramification Group

5 Higher Ramification Groups

6 Ramification Theory in the Discrete Case

Chapter 5 THE DIFFERENT

1 The Inverse Different .

2 Complementary Bases

3 Fields with Separable Residue Class Field

4 The Ramification Groups of a Subfield

Part l l Local Class Field Theory

Chapter 6 PREPARATIONS FOR LOCAL CLASS FIELD THEORY

1 Galois Theory for Infinite Extensions

3 Galois Cohomology Theory

Chapter 7 THE FIRST AND SECOND INEQUALITIES

2 Unramified Extensions

3 The First Inequality .

4 The Second Inequality: A Reduction Step

5 The Second Inequality Concluded

Chapter 8

THE NORM RESIDUE SYMBOL

1 The Temporary Symbol (c, K I KIT)

2 Choice of a Standard Generator c

3 The Norm Residue Symbol for Finite Extensions

Chapter 9 THE EXISTENCE THEOREM

1 Introduction

2 The Infinite Product Space I

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 10 APPLICATIONS AND ILLUSTRATIONS

1 Fields with Perfect Residue Class Field

2 The Norm Residue Symbol for Certain Power Series

3 Differentials in an Arbitrary Power Series Field

4 The Conductor and Different for Cyclic p-Extensions

5 The Rational p-adic Field .

6 Computation of the Index ( a : an)

Part I I I Product Formula and Function Fields in One Variable

Chapter 11

1 The Radical of a Ring 2 1 5

2 Kronecker Products of Spaces and Rings 216

3 Composite Extensions 2 1 8

4 Extension of the Valuation of a Non-Complete Field 223

Chapter 12 CHARACTERIZATION OF FIELDS BY THE PRODUCT FORMULA

2 Upper Bound for the Order of a Parallelotope 227

3 Description of all PF-Fields 230

4 Finite Extensions of PF-Fields 235

xii CONTENTS

Chapter 13 DIFFERENTIALS IN PF-FIELDS

1 Valuation Vectors Ideles and Divisors . 2 3 8

2 Valuation Vectors in an Extension Field 241

3 Some Results on Vector Spaces 2 4 4

4 Differentials in the Rational Subfield of a PF-Field 245

5 Differentials in a PF-Field 2 5 1

Chapter 14

THE RIEMANN-ROCH THEOREM

1 Parallelotopes in a Function Field 260

Chapter 15 CONSTANT FIELD EXTENSIONS

1 The Effective Degree . 2 7 1

2 Divisors in an Extension Field 2 7 8

3 Finite Algebraic Constant Field Extensions 279

4 The Genus in a Purely Transcendental Constant Field

5 The Genus in an Arbitrary Constant Field Extension 287

Chapter 16 APPLICATIONS OF THE RIEMANN-ROCH THEOREM

1 Places and Valuation Rings 293

3 Differentials and Derivatives in Function Fields 324

PART ONE

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

Theorem 1: Let I 1, and I 1, be equivalent functions, and suppose 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

3

4 1 VALUATIONS OF A FIELD 2 THE TOPOLOGY INDUCED BY A VALUATION

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

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

Set 1 a 1 , = 1 c l,Y, where y is a non-negative real number If

m/n > y , then 1 a 1 , < 1 c llmln, whence 1 an/cm 1 , < 1 Then

I an/cm 1 , < 1, from which we deduce that I a 1 , < I c I,mln Simi-

larly, if m,n < y , then I a 1 , > 1 c ImJn I t follows that I a 1, = I c Jk

Now, clearly,

1% l a I, log I a I ,

Y = l o g J c J , = l o g ~

This proves the theorem, with

I n 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

"Triangular Inequality" condition, viz., 1 a + b 1 < 1 a 1 + I b 1

T h e connection between this condition and ours is given by

Theorem 3: Every valuation is equivalent to a valuation for

which the triangular inequality holds

Proof (1) When the constant c = 2, we shall show that the

triangular inequality holds for the valuation itself

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

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

2 The Topology Induced by a Valuation

Let I I be a function satisfying the axioms (1) and ( 2 ) for valua- tions I n terms of this function we may define a topology in k by

6 1 VALUATIONS OF A FIELD

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

I t 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

Theorem 4: T h e topology induced by 1 I is HausdorfT if

and only if axiom (3) is satisfied

Proof: (1) If the topology is HausdorfT, there exist neighbor-

hoods 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

y = - xi(1 + x); then

hence

i.e I 1 + x / < llb We conclude, therefore, that if I x I < 1, then

(2) The converse is obvious if we replace I / by the equiva-

lent function for which the triangular inequality holds

I t should be remarked that the field operations are continuous

in the topology induced on k by a valuation

3 Classification of Valuations

If the constant c of axiom (3) can be chosen to be 1, i.e if

I x 1 < 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 valua-

tions we obtain a sharpening of the triangular inequality:

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,

8 1 VALUATIONS OF A FIELD 4 THE APPROXIMATION THEOREM

Proof: (1) The necessity of the condition is obvious, for if the

valuation is non-archimedean, then

(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

< D(I a In + I a In-l I b ( 4- t

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

Hence

\ a + b I < Y D ( n + l ) m a x ( I a Letting n e c o , we have the desired result

Corollary: A valuation of a field of characteristic p > 0 is

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 non-

archimedean on k

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

Theorem 7: Let I I,, a * , 1 1, be a finite number of inequiva-

lent 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)

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, a * , n - I) Let c be an ele-

ment such that I c 1 , > 1 and I c 1, < 1 We have two cases to consider:

Case I : I b 1, < 1 Consider the sequence a, = cbr Then

( a , / , = IcI, I b Ilr > 1, while I a,I, = I cI,I b Inr < 1; for suf- ficiently large r, ( a , 1, = I c 1 , ( b Ivr < 1 (v = 2, , n - 1) Thus

a, is a suitable element, and the theorem is proved in this case Case 2: ( b 1, > 1 Here we consider the sequence

cb,

1 + b"

10 1 VALUATIONS OF A FIELD 5 EXAMPLES

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

Corollary: With the conditions of the theorem, there exists

an element a which is close to 1 in I 1, and close to 0 in I 1, (v = 2,

- a , n - 1)

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 require-

ments for large enough values of r

Theorem 8: (The Approximation Theorem): Let I I,, .-., I 1,

be a finite number of non-trivial inequivalent valuations Given

any E > 0, and any elements a, (v = 1, . , n), there exists an

element a such that I a - a, 1, < E

Proof We can find elements b, (i = 1, - a , n) close to 1 in

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

upon it Consider the Cartesian product (k), x (k), x x (k),

The elements (a, 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 impos-

sibility of finding a non-trivial relation of the type

with real constants c,

5 Examples

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

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

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:

log m

(0 < a, < n; nT < m, i r r < -)

log n 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

There are now two cases to consider

Case 1: In1 > 1 for alln > 1 Then

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

12 1 VALUATIONS OF A FIELD 5 EXAMPLES 13

where a is a positive real number I t follows that

so that 1 1 is in this case equivalent to the ordinary "absolute value",

I x I = max (x, - x)

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 I t 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 non-

zero rational number, and write it as x = p~y, where the numerator

and denominator are prime to p Then I x I = cv

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 poly-

nomials 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

Case I : 1x1 > 1 Then if

c, # 0, we have

Conversely, if we select a number c > 1 and set

our conditions for functions on o are easily verified Hence this

function yields a valuation of k described as follows: let

a = f(x)/g(x), and define

deg a = deg f (x) - deg g(x)

Then [ a I = cdega Obviously the different choices of c lead only

to equivalent valuations

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 I f(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)Iv g ( 4 , (P(x), g@)) = 1, then

I f ( 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

(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 are prime to p, then we write I a ID = ( 1 1 ~ ) ~ T h e exponent v

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

(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

14 1 VALUATIONS OF A FIELD 5 EXAMPLES 15

(b) p = p , , so that I a I = @pa where c > 1 We choose

c = lld, and so define I a = d-dega

I n all cases we have made a definite choice of 1 a 1, in the equiv-

alence 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

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

If X, =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

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,) T h e

valuation I f(x) lx-xo as defined previously is now the valuation

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

I t was in analogy to this situation, that we introduced in the case

k = R, the field of rational numbers, the "infinite prime" and asso- ciated it with the ordinary absolute value

We now prove a theorem which establishes a relation between the normal valuations at all primes p:

Theorem 9: In both cases, k = R and k = F(x), the product

For q E R,

For q E k(x),

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

But I q 1, I q 1, = 1 by the theorem; hence

16 1 VALUATIONS OF A FIELD

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

+(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: For any valuation, we may take

c = m = ( I l I , ( 2 0 *

Proof (1) When the valuation is non-archimedean,

c = 1 = 1 1 1 > , 1 2 )

(2) When the valuation is archimedean, k must have charac-

teristic zero; hence k contains R, the field of rational numbers

T h e 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

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

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

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:

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

N in P; hence the residue class ring PIN is a field 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 T o justify this definition we must prove

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

1 VALUATIONS OF A FIELD

(b) that if {a,) E {b,) mod N, then lim I a, I = lim I b, I ,

(c) that the valuation axioms are satisfied

T h e 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), - a ) 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

I a - - a ; I = 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

that limV,,Ia,' /3I = O Hence lim,,,la,-PI = 0 , i.e

/3 = lim a,, and so k is complete

We now agree to identify the elements of k' with the corre-

sponding 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 valua-

tion corresponding to a finite prime p ("the completion at p") is

called the jield of p-adic numbers

CHAPTER TWO

Complete Fields

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

(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

20 2 COMPLETE FIELDS 2 EXTENSION OF THE VALUATION 2

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

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 < llv Then

and

By the induction hypothesis, the norm 1 1 1 1 on the (n - 1)-

dimensional subspace ( w , , . induces the same topology

as the special norm I I 1 lo That is,

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

if v is large enough Hence

i.e I I y I I is smaller than any chosen 7: I I y I( = 0, and so y = 0

In other words,

z;Wl + + z,,-lwn-l + wn = 0,

which contradicts the linear independence of the basis elements

ol , , 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

I I a 11, < E Thus the topology induced by norm I I I I is stronger than that induced by the special norm 1 I 11,

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

Theorem 2: 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

22 2 COMPLETE FIELDS 2 EXTENSION OF THE VALUATION 23

Thus E is complete Now let a be an element of E for which

I a I < 1; then I or Iv-+ 0; hence {av} is a null-sequence Thus if

each sequence ( x * ( ~ ) ) is a null-sequence in k

The norm of a, N(a), is a homogeneous polynomial in x, , -, x,

Hence

Hence we have proved that1 a ( < 1 3 1 N(a) 1 < 1; similarly

wecanobtain 1 a 1 > 1 + 1 N(a) 1 > 1 Thus

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

n = deg(E I k)

Then

hence I /3 I = 1 Therefore

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 esta-

blish the possibility of extending the valuation, it will be sufficient

to show that f(a) = d N(a) I coincides with I a I for a E k, and

satisfies the valuation axioms for a E E Certainly if a E k,

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

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

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/a2 I < 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

is a Cauchy sequence, for

Since k is complete, this sequence has a limit j E k Then

j2 + 1 = lim v+m x,Z + 1 = 0,

contradicting our hypothesis that a $ k This completes the proof

24 2 COMPLETE FIELDS 3 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 valua-

tion Then k has characteristic zero, and so contains a subfield R

isomorphic to the rational numbers The only archimedean valua-

tions 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, satis-

fying the following conditions:

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,, t 3 , t4 as midpoints of the sides, we define

an operator Lo by

It is easily verified that L is a linear homogeneous functional, such that L(x) = L(l) = 0, and hence vanishing on every linear func- tion

First we show that l l x is continuous for x = /3 # 0 (B E E) Let4 E Eand II 4 II < I1 B-' 11-l; then 11 518 11 < ll 4 1 1 II B-I II < 1

Since 1 ( (f/,9)v 1 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

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 = p

26 2 COMPLETE FIELDS 3 ARCHIMEDEAN CASE 27

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

the function l / ( z - a) approaches zero as I x I +co The function

is continuous for every z on the Gauss sphere, and hence is boun-

ded: say 11 l / ( x - a) 1 1 < M

We have now

1 - z + (2c - a ) ( 2 - c)"

LQ = L~ ( C - a)z -I- ( z - a ) ( c - a)z

since LQ vanishes on linear functions

Thus

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 - ZJf(3V)

contour

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

Using the estimate for L Q v [ l / ( z - a)] we find

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

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

28 2 COMPLETE FIELDS 4 THE NON-ARCHIMEDEAN CASE 29

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

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

T h e axioms (1) and (3) of Chapter I can be verified immediately

T o verify axiom (2) we notice first that

since

a,b, xk < rnax ( a,x" max I bjxj 1

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

We see at once that the last three products are smaller in valuation

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

T h e 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-archi-

medean property we obtain

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

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

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

(3) there exist polynomials A(x), B(x), C(x) and an element

d E k such that

then we can construct polynomials @(x), Y(x) E k[x] such that

Proof: As a preliminary step, consider the process of dividing

a polynomial

g(x) = b, + b,x + + bmxm

by

+(x) = a, + alx + + a,xn, where by hypothesis, I $(x) I = I anxn I The first stage in the divi- sion process consists in writing

Now

Thus we have, in fact, defined an extended valuation

30 2 COMPLETE FIELDS 4 THE NON-ARCHIMEDEAN CASE 3 1

hence we have I g l ( 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

We see that this yields

using the given bounds for I B ( x ) I, I $(x) I, I C ( x ) 1 Since I d I < 1,

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,

Referring now to our preliminary remark about division processes,

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

32 2 COMPLETE FIELDS 4 THE NON-ARCHIMEDEAN CASE 33

41(~), i,bl(x) may be repeated indefinitely to obtain approximations

which are increasingly accurate

x < 1 by the conditions of the theorem Thus $,(x) #,(x) is a better

approximation than $(x) i,b(x)

using the results obtained above

34 2 COMPLETE FIELDS

and

deg ($,(x) *,(x)) < m= {degf(x), deg h(x)}

Let

@(x) = lim {$,(x)) and Y(x) = lim {+,(x));

these limit functions are polynomials, and

deg @(x) = deg $(x)

Finally,

f (x) - O(X) Y(x) = lim {hv(x)} = 0;

thus f(x) = @(x) Y(x) Now we have only to notice that

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

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

form:

Theorem 5a: Let f(x) be a polynomial in k[x], and let the

valuation in k[x] be the special valuation just described If

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

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

(3) there exist polynomials A(x), B(x), C(x) and an element

d E k such that

f (x) = @(x) Y(x) and deg @(x) = deg $(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

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 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: x2 - a = (x - b) (x + b) + (b2 - a); h(x) = b2 - a;

and

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

Hence all the conditions of Hensel's Lemma are satisfied, and we

have a factorization f(x) = g,(x)g,(x), where deg (g,(x)) = i

Corollary: If ] f(x) 1 = 1, and f(x) is irreducible, with I a, 1 < 1,

then I ai 1 < 1 for all i > 0

Consider now

this is irreducible if f(x) is irreducible By the corollary just stated, if

I a, 1 < 1, then / a$ / < 1 for i < n Hence if 1 f(x) 1 = 1 and f(x)

is irreducible, then for 1 < i < n - 1 , I ai I < max (I a, I , I a, I),

and the equality sign can hold only if j a, I = or, I

This enables us to complete the proof that we can extend the

valuation Let us recall that all that remains to be proved is that if

or E E, then

it will be sufficient to prove the assertion for Nk(cc),k(a) Let f(x) = I r r (a, k, x), the irreducible monic polynomial in k[x] of which a is a root, be a, + a,-,x + - + xn (ai E k) Since

a, = f N(a), we see that - &

using the corollary to Hensel's Lemma

Now

f(x - 1) = Irr (a + 1, k, x),

whence

I N(1 + ff) I < 1

This completes the proof of

Theorem 7: Let k be a field complete under:a non-archime- dean valuation I 1; let E be an extension of k of degree n Then there is a unique extension of I I to E defined by

E is complete in this extended valuation

5 Newton's Polygon Let k be a complete non-archimedean field, and consider the polynomial

f(x) = a, + alx + - + anxn E k[x]

Let y be an element of some extension field of k, and sup- pose 1 y I is known Let c be a fixed number > 1, and define ord or = - log, 1 or I when or # 0; when a = 0 we write ord a = +a We shall now show how to estimate I f(y) I

We map the term a,xv of f(x) on the point (v, ord a,) in the Cartesian plane; we call the set of points so obtained the Newton

38 2 COMPLETE FIELDS

diagram The convex closure of the Newton diagram is the Newton

polygon of the polynomial (see fig 1)

Now

ord a,yv = ord av + v ord y

Thus the point in the Newton diagram corresponding to a#

lies on the straight line 1,;

y + x ord y = ord avyu

FIG 1 The absence of a point of the diagram for x = 2 means that the term

in xa is missing; i.e ord a, = m

with slope - ord y Now

o the intercept cut off on the y-axis by lV1 is less than that cut off

by lVz

Thus if I aNyN I = max, I a,yv 1, then 1, cuts off the minimum intercept on the y-axis; thus 1, is the lower line of support of the Newton polygon with slope - ord y Thus to find the maximum absolute value of the terms a,yv we draw this line of support, and measure its intercept r] on the y-axis Then max I avyv 1 = c-q

If only one vertex of the polygon lies on the line of support, then only one term avyv attains the maximum absolute value; hence

we have

If(y) ( = max I avyv I = c-"

If, on the other hand, the line of support contains more than one vertex (in which case it is a side of the polygon), then there are several maximal terms, and all we can say is that

(See figure 2.)

We shall now find the absolute values of the roots of

Let y be a root of f(x): f(y) = 0 Now 0 = I f(y) I < rnax ( avyu I , and

if there is only one term with maximum absolute value Hence if y

is a root there must be at least two terms avyv with maximum absolute value T h e points in the Newton diagram corresponding

to these terms must therefore lie on the line of support of the New- ton polygon with slope - ord y Hence the points are vertices of the Newton polygon and the line of support is a side Hence we have established the preliminary result that if y is a root of f(x) then ord y must be the slope of one of the sides of the Newton polygon of f(x)

2 COMPLETE FIELDS

FIG 2 Th: dotted lines have slope -ord f i = 4; the solid lines have slope

-ord y, = 0

Let us now consider one of the sides of the polygon, say I;

let its slope be -p We introduce the valuation of k [ x ] induced by

setting I x I = C-p, i.e

1 f(x) I = max I a , , ~ - ~ ' I

The vertices of the polygon which lie on I correspond to terms

a,xv which have the maximal absolute value in this valuation

Let the last vertex on I be that which corresponds to aixi We

define

~ ( x ) = a o + a , x + - - + a , x i ; + ( x ) = l ; f(x) - $(x) $(x) = h(x) = ~ , + ~ x ~ + l + + a,xn

Then

A(x) = C ( x ) = 0, and B(x) = d = 1

The conditions of Hensel's Lemma are satisfied since

to prove the last statement we have only to notice that

since we included in $ ( x ) all the terms with maximum absolute value; finally, $ ( x ) has absolute value equal to that of its highest term aixi Hensel's Lemma yields an exact factorization

f ( x ) = $,(x) +,(x) where $,(x) is a polynomial of degree i; hence

+,(x) is a polynomial of degree n - i From the last part of Hensel's Lemma we have

and also

I*(x) -*o(x)I < Id1 = 1-

Thus +,(x) is dominated by its constant term, 1 We notice that if

a polynomial is irreducible its Newton polygon must be a straight line; this condition, however, is not sufficient

Let us now examine the roots of $,(x), which are, of course, also roots of f ( x ) Since

42 2 COMPLETE FIELDS

the Newton polygon of +,(x) cannot lie below the side of the

Newton polygon of f(x) which we are considering; and since, by

Hensel's Lemma, +,(x) and +(x) have the same highest term, the

Newton polygon of 4,(x) must terminate at the point representing

a,xi (see figure 3) By the same reasoning as was used above we

find that if y' is a root of +,(x) then - ord y' is the slope of one

of the sides of the Newton polygon of +,(x) All these sides have

slopes not greater than the slope of 1 Hence if y' is a root of +,(x),

ord y' 2 p

We examine also the roots of #,(x) Since +,(x) is dominated by

its constant term in the valuation induced by 1, its Newton polygon

has its first vertex at the origin The origin is the only vertex of the

polygon on the line of support with slope - p, and all the sides

FIG 3 The chosen side 1 is the third side in fig 1 The Newton Polygon

of 4,(x) must: be situated like ABCD

FIG 4 The Newton Polygon of $,(x) must be situated like ODE

of the polygon have a greater slope than that of 1 Hence, by the same arguments as before, if y" is a root of #,(x), ord y" < p

Now let the sides of the Newton polygon of f(x) be I,, I,, . ,

with slopes - p, , - p, , (pl > p, > - - a ) Suppose 1, joins the points of the Newton diagram corresponding to the (iv-,)-th and (iv)-th terms of f(x) Then we have just seen how to construct polynomials +,(x) of degree iv , whose roots are all the roots yp(v)

of f(x) for which ord yp(v) )I p, Obviously ord yp(,) > pv+l , SO

that yp(") is also a root of +,+,(x); hence +,(x) divides +,,(x) We see also that the roots y of f(x) for which ord y = pv+, are those which are roots of +,,(x) but not of +,(x)

6 The Algebraic Closure of a Complete Field

Let k be a complete non-archimedean field, and let C be its algebraic closure Then we extend the valuation of k to C by defining I a I for a E C to be I a I as defined previously in the finite

44 2 COMPLETE FIELDS

extension k(a) The verification that this is in fact a valuation for C

is left to the reader; it should be remarked that the verification is

actually carried out in subfields of K which are finite extensions

of k C is not necessarily complete under this extended valuation

For instance the algebraic closure of the field of dyadic numbers

does not contain the element

The valuation induces a metric in C, and since the valuation is

non-archimedean, we have the stronger form of the triangular

inequality: I a - /3 1 < max {I a 1, 1 /3 I) Spaces in which this

inequality holds are called by Krasner ultrametric spaces Consider

a triangle in such a space, with sides a, b, c Let a = max (a, b, c);

then, since a < max (b, c) = b, say, we have a = b, and c < a

Thus every triangle is hosceles, and has its base at most equal

to the equal sides The geometry of circles in such spaces is also

rather unusual For example, if we define a circle of center a and

radius r to consist of those points x such that I x - a I < r, it is

easily seen that every point inside the circle is a center We now

use this ultrametric geometry to prove:

Theorem 8: Let a E C be separable over k, and let

r = min I u(a) - CL ( ,

a# 1

where the a are the isomorphic maps of k(a) Let /3 be a point, i.e

an element of K, inside the circle with center a and radius r

Then k(a) C k(/3)

Proof: Take k(/3) as the new ground field; then

f(x) = Irr (a, k, x)

is separable over k(/3) If $(x) = Irr (a, k(/3), x), then $(x) I f(x)

Let o be any isomorphic map of k(a, /3) I k(/3) Since

and conjugate elements have the same absolute value (they have

the same norm) we deduce that 1 /3 - o(a) 1 = I - a 1 < r

Consider the triangle formed by /3, a, u(a); using the ultrametric

property we have as above I a - @(a) I < I /3 - a I < r, i.e o(a) lies inside the circle Hence o(a) = a, and so, since a is separable, the degree [k(a, /3) : k(@] is equal to 1, and k(a) C k(/3)

Let f(x) be a polynomial in k[x] with highest coefficient 1 In C[x], we have f(x) = (x - a,) (x - a,) Suppose that in the valuation of k[x] induced by taking I x I = 1, we have I f(x) I < A, where A 1 Then if a E C, and I a I > A, we see that an is the dominant term in

f(a) = Clr + upn-I + + a,

Hence a cannot be a root of f(x) = 0 Hence if I f(x) I < A and

a, , -, an are the roots of f(x) = 0, then I a, I < A

Consider now two monic polynomials f(x), g(x) of the same degree, n, such that I f(x) - g(x) 1 < E Let /3 be a root of g(x),

a, , + - - , an the roots of f(x) Then

where A is the upper bound of the absolute values of the coefficients, and hence of the roots, of f(x) and g(x) Hence

and so one of the roots ai , say a, must satisfy the relation

IB-.,I < A &

Thus by suitable choice of E, each root /3 of g(x) may be brought as close as we wish to some root ol, of f(x) Similarly by interchanging the roles of f(x) and g(x), we may bring each root ai of f(x) as close

as we wish to some root of g(x) Let us now assume that

E has been chosen such that every /3 is closer to some a( than min / ai - O L ~ 1; ai # O I ~ in this way the /3's are split into sets

"belonging" to the various q

Suppose for the moment that f(x) is irreducible and separable Then, since I /3 - ai I < min I ai - olj I, the preceding theorem gives k(/3) 3 k(a); but f(x) and g(x) are of the same degree, whence

Theorem 9: If f(x) is a separable, irreducible monic polyno- mial of degree n, and if g(x) is any monic polynomial of degree n

such that I f(x) - g(x) I is sufficiently small, then f(x) and g(x)

46 2 COMPLETE FIELDS 7 CONVERGENT POWER SERIES 47

generate the same field and g(x) is also irreducible and separable

If f(x) is not separable, let its factorization in C[x] be

with distinct % I n this case we can establish the following result:

Theorem 10: If g(x) is sufficiently close to f(x), then the number

of roots ,tId of g(x) (counted in their multiplicity) which belong to

ffl 1s V l

Proof: If the theorem is false, we can construct a Cauchy

sequence of polynomials with f(x) as limit for which we do not have

vc roots near oli From this sequence we can extract a subsequence

of polynomials for which we have exactly pi roots near cq (pi f v$

for some i) Since k is complete, the limit of this sequence of poly-

nomials is f(x), and the limits of the sets of roots near ai are the ari

Hence we have

(X - alp (X - a p (X - a J r = f((x)

= (x - al)"l (x - ~ r ~(x - ) a,)"' ~ ~This contradicts the unique factorization in C[x], so our theorem

is proved

In a similar manner we can prove the result

Theorem 11: If f(x) is irreducible, then any polynomial

sufficiently close to f(x) is also irreducible

Proof: If the theorem is false, we can construct a Cauchy

sequence of reducible polynomials, with f(x) as limit From this

sequence we can extract a subsequence {g,(x)): g,(x) = h,(x) m,(x)

for which the polynomials h,(x) have the same degree, and have

their roots in the same proximity to the roots of f(x) Then the

sequence {h,(x)) tends to a limit in k[x], whose roots are the roots

of f(x) This contradicts the irreducibility of f(x)

Now although k is complete, its algebraic closure C need not

be complete; the completion of C, c , is of course complete, but

we can prove more:

Theorem 12: f? is algebraically closed

Proof: We must consider the separable and inseparable polynomials of C[x] separately

(1) Let f(x) be a separable irreducible polynomial in c[x] The valuation of k can be extended to a valuation for the roots of f(x)

We can then approximate f(x) by a polynomial g(x) in C[x] suffi- ciently closely for the roots of f(x) and g(x) to generate the same field over f? But g(x) does not generate any extension of I?; hence the roots of f(x) lie in C

(2) If the characteristic of k is zero, there are no inseparable polynomials in c[x] Let the characteristic be p # 0; if ar is an element of C, a is defined by a Cauchy sequence {a,) in C But if {a,) is a Cauchy sequence, so is {a,llp), and this sequence defines orllp, which is therefore in c Hence there are no proper inseparable extensions of c

7 Convergent Power Series

Let k be a complete field with a non-trivial valuation / I Con- vergence of series is defined in k in the natural way: Zzm a, is said to converge to the sum a if for every given E > 0 we have

I Zil, a, - a I < E for all sufficiently large n The properties of the ordinary absolute value which are used in the discussion of real

or complex series are shared by all valuations I I Hence the argu- ments of the classical theory may be applied unchanged to the case

of series in k In particular we can prove the Cauchy criterion for convergence, and its corollaries:

1 The terms of a convergent series are bounded in absolute value

2 If a series is absolutely convergent (i.e if Emm I av 1 converges

in the reals) then the series is convergent

Let E be the field of all formal power series with coefficients in k:

E consists of all formal expressions f(x) = Z z a,xv, with a, E k, where only a finite number of the terms with negative index v are non-zero We define a valuation I 1, in E such that I x 1, < 1, and

I 1, is trivial on k If an element is written in the form E: a,xv with avo # 0, then 1 E t a,xv 1, = 1 xvo 1, = (1 x I,).o This

48 2 COMPLETE FIELDS 7 CONVERGENT POWER SERIES

valuation ( I, on E and the valuation I I on k are therefore totally

unrelated

An element f(x) = I:avxv in E is said to be convergent for the

value x = c E k (c # 0), when E a c V converges in k We shall

say simply that f(x) is convergent if it is convergent for some c # 0

Theorem 13: If f(x) is convergent for x = c # 0, then f ( x )

is convergent also for x = d E k, whenever I d 1 < I c 1

Proof: Let I c 1 = lla

Since I:a,,cv is convergent, we have 1 a,,cv I < M, whence

l a v l < M a v If [ d l < I c l , then

Since a I d I < I, the geometric series E M(a I d I)v is convergent

Thus E I a,& ( is convtirgent; since absolute convergence implies

convergence, this proves the theorem

Now let F be the set of all formal power series with coefficients

in k which converge for some value of x E k (x # 0) It is easy to

show that F is a subfield of E Notice, however, that F is not

complete in I 1,

Theorem 14: F is algebraically closed in E

Proof: Let 8 = I:zm afiv be an element of E which is algebraic

over F We have to prove that 8 lies in F

I t will be sufficient to consider separable elements 8 For if 8 is

inseparable and p # 0 is the characteristic, then 8p' is separable

for high enough values of r But 8pT = I:> av~'xvp', and the con-

vergence of this series implies the convergence of 8

Let f(y) = Irr (8, F, y); we shall show that we may confine our

attention to elements 6' for which f(y) splits into distinct linear

factors in the algebraic closure A of E Let the roots of f(y) in A

be 8, = 8, 8, , ., 8, Since the valuation I 1, of E can be extended

to A, we can find I Oi I , Let

of pl obviously implies the convergence of A,, and hence of 8,,

it follows that we need only consider elements 8 which are separable over F, and whose defining equation f(y) has roots 8, = 8,8, , -., 8,

in A with the property that 1 8, 1 ,< I x 1, and I 8, 1, > 1 for i > 1

We may further normalize f(y) so that none of the power series appearing as coefficients have terms with negative indices, while

at least one of these series has non-zero constant term, i.e

where at least one b,, # 0

If we draw the Newton diagram of f(y), it is obvious from the form of the coefficients that none of the points in the diagram lie below the x-axis, while at least one of the points lies on the x-axis; further, since f(y) is irreducible, its Newton polygon starts on the y-axis The first side of the Newton polygon of f(y) corresponds to the roots y of f(y) for which ord y is greatest, and it has slope - ord y Only 8, = 8 has the maximum ordinal; hence the first side of the polygon joins the first and second points in the Newton diagram, and its slope is - ord 8, < 0 The other sides

of the polygon correspond to the remaining roots 8,, and have slopes - ord Bi > 0 Hence the Newton diagram has the form shown:

2 COMPLETE FIELDS

Thus

where blo # 0 Since y = 8 is a root of f(y), we have

whence

But we know that f(y) has a root given by

From these two formulas we obtain a recursion relation:

These polynomials tL, have two important properties:

(a) They are universal; i.e the coefficients do not depend on the particular ground field k

(b) All their coefficients are positive

Now since the series EcP,xV are convergent, we have

I ~ , , ~ a , , ~ I < MP for all q, E k having I or, I < some fixed lla, Since the number of series occurring as coefficients in f(y) is finite, we can write I c,, I < Mav, where

M = max (M,) and a = max (a$

We now go over to the field of real numbers, ko , and the field

of convergent power series Fo over k, Let 4 be a root of the equation

where the sum on the right is infinite:

4, = 2; a,xY We shall now show that all the coefficients orv are positive, and I a, I < a, We proceed by induction; a, is easily

52 2 COMPLETE FIELDS

shown to be positive and / a , / < al; hence we assume a, 2 0 and

I a, I < a, for i < m Since 4, satisfies the equation (2), ~ t s coeffi-

cients satisfy

am+l = $rn(al, .", a m ; Mav)

Since only positive signs occur in $ m , we have a , 2 0 For

a,,, we already have expression (1) and hence

Again using the fact that only positive signs occur in #m , we have

Hence I a , I < a,, where the a, are coefficients of a convergent

power series Thus 8 = E a 3 ' is convergent

This completes the proof of the theorem

CHAPTER THREE

e, f and n

1 The Ramification and Residue Class Degree

The value group of a field under a valuation is the group of non-zero real numbers which occur as values of the field elements From now on, unless specific mention is made to the contrary,

we shall be dealing with non-archimedean valuations For these

since I a I # 0 and I a,, - a I can be made as small as we please,

in particular less than I a I by choosing p large enough Thus

I a I = I a, 1 for large enough p This proves the theorem Now let k be a non-archimedean field, not necessarily complete; and let E be a finite extension of k If we can extend the valuation

of k to E, we may consider the value group BE of E Then the value group B, of k is a subgroup, and we call the index

e = (BE : B,) the ramiJication of this extended valuation

Consider the set of elements a E k such that I a I 5 1 I t is easily shown that this set is a ring; we shall call it the ring of integers

and denote it by o The set of elements a E o such that 1 a I < 1

53

1 THE RAMIFICATION AND RESIDUE CLASS DEGREE 55

forms a maximal ideal p of o; the proof that p is an ideal is obvious

T o prove it is maximal, suppose that there is an ideal a # p such

that p C a C o Then there exists an element a E a which is not

in p; that is I a 1 = 1 Hence if /3 E 0, /3/a E o also, and

= a (/3/a) E a That is, a = o Hence p is a maximal ideal and the

residue class ring o/p is a field, which we denote by li If & is the

completion of k, we can construct a ring of integers 5, a maximal

ideal p, and a residue class ring 5/ii = fi We have the result

expressed by:

Theorem 2: There is a natural isomorphism between ?t and fi

Proof: We have seen that if a E 6, then a = lim a, E o and

I a I = I a, I for v large enough Thus 5 is the limit of elements

of o; i.e 6 is the closure of 0 Similarly p is the closure of p

Consider the mapping of o/p into ii/p given by a + p -+ a + 6

This mapping is certainly well-delined, and is an "onto" mapping

since, given any a E 5, we can find an a E o such that I a - a I < 1 ;

then a + f~ = a + @, which is the image of a + p The mapping

is (1, 1) since if a, b E o and a - b mod j3 we have I a - b I < 1;

hence a r b mod p I t is easily verified that the mapping is homo-

morphic; hence our theorem is completed

From now on we shall identify the residue class fields K, fZ under

this isomorphism

Now let E be an extension field of k, with ring of integers L) and

prime ideal p Then L) 3 o = L) n k and Q 3 p = Q n k

Theorem 3: There is a natural isomorphism of the residue

class field ?t onto a subfield of the residue class field E

Proof: Consider the mapping a + p -+ a f Q ( a E k) This

is well-defined since a + p + 3 = a + Q; it is easily seen to be a

homomorphism of 6 into E Finally, the mapping is (I, 1) onto the

image set, for

We shall now identify ?t with its image under this isomorphism

and so consider E as an extension field of A We denote the degree

of this extension by [E : K] = f

Let w1 , w , , - , or be representatives of residue classes of E

which are linearly independent with respect to K ; that is, if there exist elements c, , c, , c, in a such that c1wl + + c,w,

lies in Q, then all the ci lie in p Consider the linear combination

c,w, + + c,w, , with ci E o; if one of the ci , say c, , lies in o but not in p, then c,w, + - + c,w, + 0 mod '$3, and hence

Now consider the linear combination dlwl + + d p , , with

di E k, and suppose dl has the largest absolute value among the di

Then

Hence if wl , a , w , are linearly independent with respect to K, then

I d,wl + + d,w, / = max v I d, (

Let now n1 , n 2 , ns be elements of E such that I n, I, - , I .rr, I

are representatives of different cosets of We shall use these

wi , 9 to prove the important result of

Theorem 4: If E is a finite extension of k of degree n, then

ef < n

Proof: Our first contention is that

Certainly

and since the I ni 1 represent different cosets of the

I Z, c,w,ri I are all different Hence

as required

2 T H E DISCRETE CASE 57

I t follows that the elements w , ~ , are linearly independent, since

Hence rs < deg ( E / k) Thus if deg (E I k) = n is finite, then e

and f are also finite, and < n

The equality ef = n will be proved in the next section for a very

important subclass of the complete fields, which includes the cases

of algebraic number theory and function theory Nevertheless,

the equality does not always hold even for complete fields, as is

shown by the following example

Let R, be the field of 2-adic numbers, and let k be the completion

of the field R,(& Y2, $9, a + ) Then deg (A(-) : k) = 2,

but the corresponding e and f are both 1 This is

fact that we may write

This series is not convergent, but if we denote by

the first n terms, and by ui the i-th term, we have

caused by the

s, the sum of

2 The Discrete Case

We have already introduced the function ord a = - log, I a I,

c > 1 The set of values taken up by ord a for a # 0, a E k forms

an additive group of real numbers; such a group can consist only of

numbers which are everywhere dense on the real axis or which

are situated at equal distances from each other on the axis Hence

we see that the value group, which is a multiplicative group of

positive real numbers, must be either everywhere dense, or else

an infinite cyclic group I n this latter case the valuation is said to

be discrete

I n the discrete case, let T be an element of k such that I T I

takes the maximal value < 1 Then the value group consists of the

numbers I T Iv; given any non-zero element a E k, there is a positive

or negative integer (or zero) v such that I OI 1 = 1 T 1' Then

I a/+ I = 1 = I r v / a I; hence a/.rrV and its inverse are both elements

of D, i.e a / r V is a unit E of o For every a # 0 we have a factorization

a = rVe, where I E I = 1; hence there is only one prime, namely T Since k and its completion & have the same value group, the same element T can be taken as prime for &; thus if a complete field & is the completion of a subfield k, the prime 'for & may be chosen

in k

Now let k be a complete field under a discrete valuation If

a E k can be written as a = rive, we shall take ord a = v; in other words, we select c = 111 T I We now suppose that for every positive and negative ordinal v an element n, has been selected such that I n, I = I T Iv-obviously T, = rV would suffice, but we shall find it useful to use other elements Suppose further that for each element in h we have selected a representative c in o, the representative of the zero residue class being zero Then we prove

Theorem 5: Let k be a complete field with discrete valuation Every a E k can be written in the form a = C c , ~ , , where

n = ord a and c, f 0 mod p

Proof: When a = 0, there is nothing to prove, so we suppose

a # 0 Since ord a = n, we have I a I = I T, I, hence a/r, is a unit E Its residue class modulo p is represented by c, E E mod p

Thus I e - c n I < 1, whence I E T , - C , ~ ~ , / < \ % I ; i.e

a = E T ~ = C,T, + a' where I a' I < I T, I We may repeat the procedure with a', and so on, obtaining at the m-th stage

where I a ( m ) I < I rm I; thus a ( m ) + 0 as m +a This proves the theorem If the representatives ci and the T, are chosen once for all, then the series representation of a is unique

I t is important to notice that the c, and T, are chosen from the same field k; thus the characteristic of the field containing the c,

is the characteristic of k, not of h We shall illustrate this remark

by considering the valuation induced on the rational field R by a finite prime p Any element a E R can be written as a = pu (blc) where b and c are prime to p; the ring of integers o consists of

58 3 e, f AND n 2 THE DISCRETE CASE 59

these elements a for which v 2 0, and the prime ideal p consists

of the a such that v > 0 Let R, be the completion of R under this

valuation, 5, 6 the corresponding ring and prime ideal Since

a/+? = o/p, we can choose representatives for the residue classes

in R, and in fact in o The residue classes are represented by

the elements b/c E R where b and c are prime t o p ; we can solve

the congruence ca = 1 modp in integers Then bd is a repre-

sentative of the residue class containing blc, since

Hence every residue class contains an integer, and a complete set

of representatives is given by 0, 1,2, ., p - 1 This set of elements

does not form a field since it is not closed under the field operations

of R Now every p-adic number a can be written a = E:cqv;

we see that the field of p-adic numbers, R, , has characteristic

zero I t can be shown that the field R, contains the (p - 1)-th

roots of unity; it is often convenient to choose these as representa-

tives of the residue classes

This analysis of complete fields may also be used to prove that

the field of formal power series k = F{x) over any field F is com-

plete Any element a E k can be written as a = Er c,xV, c,, E F

The ring of integers o, respectively the prime ideal p are made

up of the elements a for which n 2 0, respectively n > 0 We can

choose x = rr; and as representatives of o/p the elements of F;

hence the completion of k consists of power series in x with coef-

ficients in F; i.e k is complete

Now let k be a complete field with discrete valuation; let E

be a finite extension with degree n and ramification e The finiteness

of e shows that the extended valuation is discrete on E Let 17, .rr

be primes in E, k respectively; then the value groups are

B E = {I 17lV}, Bk = { I n - 1,) Since ( B E : 23,) = e, 1171e = I T 1;

hence .rr = E De We shall find it more convenient to represent B E

as {I .rrVncl 1) where a < v <a and 0 < p ,< e - 1

Theorem 6: If the valuation is discrete, then ef = n

Proof: Let w, , w, , - a , wf be elements of E which represent a

basis of the residue class field Ep/kp Thus the generic residue

class is represented by c p , + c202 + + cfwf, with c4 E o Then if a is an element of E, a can be written

Thus every element a E E can be expressed as a linear combination

of the ef elements w p with coefficients in the ground field k

Thus the degree of the extension E I k is at most ef: n < ef

We have already seen that for all extensions, whether the valua- tion is discrete or not, n 2 ef

Hence n = ef and our theorem is proved

Theorem 7: The elements (w$7~} (P = 1, f ; p = 0, - - * ,

e - 1) form a basis D over o

Proof: We have already seen, in the course of the last proof, that {wp17/1} form a basis

Let a be an integer of E We can write a = E,,, d,,&,lIfl, and since I w, I = 1, and the ( 17, ( are all distinct, we have

3 The General Case

I n this section we shall prove that if k is complete under an

arbitrary valuation, and if E is a finite extension of degree n, with

ramification e and residue class degree f, then ef divides n, and the

quotient 6 is a power of the characteristic of the residue class field

k, This section is added for the sake of completeness, and the

result will not be used in the sequel,

Our first step is the proof of a weaker result:

Theorem 8: The only primes which can divide e and f are

these which divide n

Proof: (1) The proof for e is simple

Since I or I = N(or)lJn, I a In lies in the value group of k The

factor group B,/B, is abelian of order e Hence e can contain only

primes dividing n

(2) T o prove the analogous result for f, we introduce the notion

of the degree, deg a, of an element a E E; by this we shall mean the

degree of the irreducible equation in k of which a is a root; hence

deg or = [k(u) : k] Similarly the degree of a residue class

6 E Ep is the degree of the irreducible congruence in k of which 6i

is a root; hence deg 2 = [K(&) : El

We shall prove that if a is an integer of E, and 6i the residue

class in which it lies, then deg & divides deg a

Let

f(x) = Irr (or, k, x) = xn + alxn-l + - + a,

Since or is an integer, ( a I < 1, and so I a, / < 1 By the corollary

to Theorem 6, Chapter 11, this implies that all the coefficients of

f(x) are integers: I a4 I < 1 Now Hensel's Lemma may be used

to show that f(x) cannot split modulo p into two factors which are

relatively prime For if f(x) = +(x) $(x) mod p , where +(x) and

+(x) are relatively prime, there exist polynomials A(x), B(x) such

such that

A(x) +(x) + B(x) $(x) = 1 mod p

3 THE GENERAL CASE 61

where ( C(x) I < 1 And since f(x) r +(x) #(x) mod p , we have f(x) = +(x) $(x) + h(x), with I h(x) I < 1 By Hensel's Lemma, this situation implies a factorization of the irreducible polynomial f(x), which is impossible

Hence if f ( x ) splits modulo p, it must do so as a power of an irreducible polynomial: f(x) r [P(x)]P But deg Z = deg P(x), and deg or = deg f(x), hence deg & divides deg or as required

Thus if & is any residue class of E, , and a E E is any representa- tive of E, then deg 2 divides deg a But since deg a = [k(a) : k] divides n, this implies that deg 2 can be divisible only by primes which divide n Let E,' be the separable part of E, The degree

of E, I E,' is some power pv of the characteristic of kp I f f = f'pv,

then E,' = A,(&), where deg & = f'; f ' is divisible by all the primes dividing f except possible the prime p Hence, by our preceding remarks, these primes must divide n

Finally, if 2 is an inseparable element, then p divides deg 15;

hence p must divide n This completes the proof

Theorem 9: If deg (E I k) = q, where q is a prime not equal

to the characteristic p of the residue class field, then ef = q Proof: Since q is not the characteristic of the residue class field,

q does not lie in the prime ideal; hence ( q 1 = 1 Furthermore,

q is not the characteristic of k, and hence E I k is separable Let

E = k(a), and let

f(x) = Irr (a, k, x) = XI + a1xq-l + - + a,

We may apply the transformation x = y + a,/q, since q is not the

characteristic; f(x) assumes the form y'l + b,yq-l + + bq

Thus we may assume at the outset that a, = 0 We remark that

If I cx I does not lie in the value group 13, , we have e 2 q, since

I or ( Q E Bk Since ef < q, we obtain e = q and f = 1 ; hence ef = q

If, on the other hand, I a I lies in Bk , we have I or ( = 1 a 1

where a E k We may write = a& and will satisfy an equation with second coefficient zero; hence we may assume without loss of generality that I or 1 = 1

I n the course of Theorem 8 we saw that f(x) either remains

3 THE GENERAL CASE 63

irreducible modulo p or splits as a power of an irreducible poly-

nomial; since q is a prime the only splitting of this kind must be

j(x) = ( x - c)q mod p We show now that this second case is not

possible: since a, - cq, mod p, c + 0; hence the second term

of (x c),, namely qcq-l, is + 0 mod p T h i s contradicts our

assumption about f(x)

T h u s n satisfies an irreducible congruence of degree q; f 2 q

Hence f = q, e = 1, and ef = q

This proves the theorem

I n order to prove that ef divides n, we shall require the

Lemma: If E 3 F 3 k, then e(E 1 k) = e(E I F) e(F 1 k), and

similarly for f, and hence for ef

T h e verification of this is left to the reader I t follows at once

that for a tower of fields with prime degree unequal to the character-

istic of the residue class field, ef = n We shall now prove the main

result

Theorem 10: ef divides n; the quotient is a power of the

characteristic of the residue class field

Proof: We shall prove the result first for the case where E is a

separable extension of k Let deg (E / k) = n = qinz, where q is a

prime unequal to the characteristic of k and jq, 7%) = I We assert

that ef = qis where (q, s) = 1

Let K be the smallest normal extension of k containing E, and let G be its Galois group; let H be the subgroup of G corresponding

to E Let Q be a q-Sylow subgroup of H, E, the corresponding field Now Q, considered as subgroup of G, may be embedded in a q-Sylow subgroup Q' of G ; the corresponding field El is a subfield

of E,

T h e degree of E, 1 E is prime to q since Q is the biggest q-group

of H Similarly the degree of El I k is prime to q, since Q' is the biggest q-group in G Now the degree of E, / El is a power of q, which must be qi, for

[E, : EJ [El : k ] = [E, : k ] = [ E : k ] [E, : El

Now between the groups Q and Q' there are intermediate groups, each of which is normal in the one preceding and each

of which has index q T h u s the extension E, I El may be split into steps, each of degree q T h e product ef for the extension

E, I El is equal to the corresponding product for each step of degree q From this we see that ef = qi But this is also the q-contribution of ef by the extension E 1 k, since the degree of

E, I E is prime to q

We obtain this result for every prime divisor of n unequal to the characteristic of A Hence ef can differ from n only by a power

of the characteristic Since ef < n, we must have efS = n where

6 is a power of the characteristic This proves the theorem for separable extensions

If E I k contains inseparable elements, let Eo be the largest separable part Then n[E : Eo] must be a power of the characteristic

of k; if this is non-zero it is also the characteristic of A We have seen that ef[E : Eo] can be divisible only by this prime Hence for the inseparable part we have again nlef = a power of the character- istic of A This completes the proof of the theorem

Corollary: If the residue class field has characteristic zero, then ef = n

We may write n = ef6 : 6 is called the defect of the extension

1 UNRAMIFIED EXTENSIONS 65

CHAPTER FOUR

Ramification Theory

1 Unramified Extensions

Let k be a complete field, C its algebraic closure Let the cor-

responding residue class fields be k, e; under the natural isomor-

phism, k may be considered as a subfield of C The canonical

image in C of an integer a in C shall be denoted by 5; that of a

polynomial $(x) in C[x] by $(x) A given polynomial $(x) in €[XI

is always the image of some polynomial $(x) of C[x]; +(x) may be

selected so that it has the same degree as $(x) If the leading

coefficient of #(x) is 1 we may assume that the leading coefficient

of +(x) is also 1 In the sequel these conventions about the degree

and leading coefficients will be tacitly assumed

Let $(x) = $(x) be an irreducible polynomial in &[XI, with

leading coefficient 1 We may factor +(x) in C[x]:

Since all the roots are integers, we may go over to €[XI, where

This shows that $(x) splits into linear factors in €[XI; hence is

algebraically closed

Since $(x) is irreducible in Qx], $(x) is irreducible in k[x], and

hence F = k(/3,) has degree n over k The residue class field P

contains the subfield k(&), which is of degree n over k; hence

f = deg (F 1 5 ) 2 n But if e is the ramification, we have ef < n

Hence e = 1, f = n, and F = K(pl) This shows that every simple

extension k(pl) of k is the residue class field of a subfield F of C,

with the same degree as K(pl) I h, and ramification 1

Since all finite extensions of & may be obtained by repeated simple extensions, we have proved

Theorem 1: Every finite extension of is the residue class field for a finite extension of k with ef = n and e = 1

We return to the case of a simple extension, and assume now that #(x) is separable Then pi # Pj for i # j, and hence

j pi - pi I = 1 for i # j Let ar be an integer of C such that

$ ( C ) = O ; say & = P I Then Iar-/3,j < 1; that is, (a /3,I

is less than the mutual distance of the & Theorem 7 of Chapter

2 shows that k(/3,) C k(a)

Let E be any subfield of C such that 3 &(PI) Then E contains

an element a such that E = PI; hence k(pl) C k(a) C E If we assume in particular that deg ( E 1 k) = deg ( E I A) and that

E = k(pl), then k(/3,) C E, but

hence E = A(/?,) Thus we have proved

Theorem 2: T o a given separable extension k ( a ) of 6, there corresponds one and only one extension Eo of k such that (a) deg (Eo I k) = deg (Eo 1 k) and (b) Eo = k(p1)

Corollary: If E is an extension of k such that E contains Eo ,

(3) E 1 k is separable

The third condition is inserted to exclude the critical behavior

of the different when inseparability occurs This difficulty does not arise in the classical case of the power series over the complex numbers, where the residue class has characteristic zero; nor in the case of number fields, where the residue class fields are finite

I n neither of these cases is any inseparability possible