Holomorphic functions, domains of holomorphy and local properties

vv AND LOCAL PROPERTIES Dedicado h minha filha L ~ NACHBIN A NORTH-HOLLAND MATHEMATICS STUDIES 1 Holomorphic Func...

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vv AND LOCAL PROPERTIES

Dedicado h minha filha L ~ NACHBIN A

NORTH-HOLLAND MATHEMATICS STUDIES

1

Holomorphic Functions, Domains of Holomorphy and Local Properties LEOPOLDO NACHBIN George Eastman Professor University of Rochester Rochester, New York, USA

Notes prepared by Richard M. Aron

i 970 NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM

. LONDON

0 NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM - 1970 All Rights Reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the. Copyright owner.

Library of Congress Catalog Card Number 78-134642 ISBN North-Holland 07204 2041 5 ISBN American Elsevier 07444 10054 7

PUBLISHERS :

NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM NORTH-HOLLAND PUBLISHING COMPANY, LTD. - LONDON SOLE DISTRIBUTORS FOR THE U.S.A. AND CANADA:

AMERICAN ELSEVIER PUBLISHING COMPANY, INC. 52 VANDERBILT AVENUE NEW YORK, N.Y. 10017

PRINTED I N THE NETHERLANDS

FOREWORD

This t e x t i s p a r t l y based on courses t h a t I taught a t t h e I n s t i t u t o de Matematica Pura e Aplicada, Rio de J a n e i r o i n 1958 and 1967, a t t h e Universidade do Recife i n 1959, and a t t h e Universidade de Sa"o Paul0 i n 1967.

It reproduces l i t e r a l l y

a course t h a t I taught a t t h e University o f Rochester during t h e F a l l of 1967. These l e c t u r e s were intended as an elementary i n t r o d u c t i o n t o s e v e r a l complex v a r i a b l e s , arranged t o lead t o c e r t a i n f e a t u r e s of t h e s u b j e c t having no c o u n t e r p a r t i n a standard Domains of holomorphy and

treatment of one complex v a r i a b l e .

l o c a l p r o p e r t i e s seem t o be e s p e c i a l l y s u i t a b l e i n giving a beginner i n t h e f i e l d a f l a v o r o f how s e v e r a l complex v a r i a b l e s d i f f e r from one complex v a r i a b l e .

Limitation i n time and t h e

d e s i r e t o present t h e m a t e r i a l i n an e x p l i c i t form had t h e i r n a t u r a l i m p l i c a t i o n i n t h e s t y l e of t h e c o u r s e .

of t h i s volume are encouraged t o use t h e materia

The r e a d e r s quoted i n

t h e Bibliography i n o r d e r t o f u r t h e r t h e i r study i n t h e d i r e c t i o n of l o c a l and g l o b a l complex a n a l y s i s as needed i n present-day mathematics and physics. Thanks a r e due t o M r . Richard M. Aron, of t h e University

of Rochester, f o r preparing t h i s t e x t . LEOPOLDO NACHBIN

Rochester December 1967 V

This Page Intentionally Left Blank

TABLE OF CONTENTS

Chapter 1: Holomorphic Functions

1 2 3 4

. . . .

Section 5

.

Sec t i o n Section Section Section

Section 6 Section 7 Section 8 Chapter 2 :

. .

Domains of Holomorphy

Section Section Section Section Section Chapter 3 :

.

............ 1 ..... 4 ........................................ 9 ................................. 12 ...........................................16 ......... 25 ................................30 ................................... 31

Ho lomorphic Functions of 1 Complex Variable Holomorphic Functions of Several Complex Variables Cauchy I n t e g r a l D i f f e r e n t i a t i o n of Holomorphic Functions and the Cauchy I n e q u a l i t i e s The Natural Topology on t h e Spaces of Holomorphic Functions Taylor S e r i e s and Unique Analytic Continuation Maximum Modulus Theorem Holomorphic Mappings

.

.

................................ 34

.

49 59

............... .......... 7 1

9 Removable S i n g u l a r i t i e s 10 Domains of Holomorphy .................................. 11. The Cartan-Thullen Theorem ............................. 12 Open S e t s of Convergence of Power S e r i e s 13 Further P r o p e r t i e s of Open S e t s of Holomorphy

.

44

Local P r o p e r t i e s

.

............................ 78 .................. 84 ................................ 94 . ........................ 104 122 Biblionraphy ............................................................. Section Section Section Section

14 15 16 17

. .

Germs of Analytic Functions The Division and Preparation Theorems The Noetherian Property Unique F a c t o r i z a t i o n P r o p e r t i e s

vii

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Chapter 1: Holomorphic Functions

Notation:

If x

=

(x~,..~,x) E Rn, then x may be regarded n j

as a map:

xj' If a = (al,...,a +

integer, then

), where each a

n la1 = al

Let U cRn, open.

+

... + an

=

Cm(U) = Cm(U,(c)

is a non-negative j order of a and a ! = al!.

=

61.

=

n c j=l

cl(u>,

=

.

O,l, ...,a

A dx axj

j'

Holomorphic Functions of 1 Complex Variable.

Let f : U+C, where U c f E

I

set of all f : U 3 C , with

continuous partial derivatives up to order m, m

df

..a n9

(c,

open,

in the real sense (i.e.

(c

=

IR2

Let

U is considered as U spiz ).

Then

Now,

z = x

functions, and

-=x-

+ iy, z

z =

id : (c 3

(c,

iy and

(where x and y are projection is the complex conjugate

HOLOMORPHIC FUNCTIONS

2

mapping). dY

-

-

dz 2idz

=

z + z 2 '

x =

So,

z - z

Y

=

T

By substitution, df =

, dx af ax

=

dz

(

dz

+ dz 2

'

+ dg )

~~

Definition 1. We shall define the operators

Clearly,

af = ax

az

+ af af'

and

f

C1(U)

Definition 2 .

41 af. = af - af ay az azw

af ax

4hf 1

words, f E C1(U)

:

1

Then f E C (U)

af az =

2

0

-

iff u,v

(= =) +

ax

+ &iZ az

df = &z az

af = az

(complex df)

0 (Cauchy-Riemann equation),

ay

f is holomorphic on U).

is holomorphic if and only if

Remark 1. Write f

=

az

= 0).

Let #(U) = [f : U - + C

@ az

and

az

is analytic or holomorphic in U

in the Cauchy-Riemann sense iff (iff

-

a (dz iidz) = +

2

=

u

6

+

In other =

ax

af . az

iv, where u : U A R, v : U-7R.

C1(U).

a on U al,

,U

c Cn, open, then we say that

f

is differentiable

in the complex sense if, for all z E U,

.. .,an E

(c)

such that

there are


6

f(zl+hl,

...,z n+hn) 11

as h+O,

f(zl,

n

..., z n) -

c ajhj

i=l

3 0

hll

where h = (hly...,hn) E

n

(2

, and

llhll

f is differentiable in the complex sense on U and

sup {lhil :i=l,...,n},

Then Goursat's Theorem says that

(or any other norm on C").

if f E C1(U)

=

if and only

af = 0, for all az

j = l,...,n. Goursat's j Theorem in dimension n is a trivial consequence of Goursat's Theorem in dimension n Remark 3 . on

=

1.

It is clear that if f : U

is holomorphic

U, then f is partially holomorphic with respect to each

variable. Conversely, there is a deep theorem of Hartogs saying that if f is partially holomorphic, then

f is holomor-

phic. This result has no analogue for real differentiability or continuity. Remark 4 .

If f : U 9 6

,U

c Cn, open,

f E C1(U),

we

define the complex differential af and the complex conjugate differential 3fy by so

af

=

c ~a az =f ~dzj j

that the real differential (written d

and

-af = xy=l

5 j

=

a + 5) df

=

af

dZ j'

+ df.

.

Then, the Cauchy-Riemann system of equations qf = 0 (j=1,.. ,n) is equivalent to the Cauchy-Riemann equation

azj -af =

0. Our pre-

7

HOLOMORPHIC F U N C T I O N S

s e n t a t i o n h e r e w i l l be i n t h e s e v e r a l complex v a r i a b l e s t y l e . I f w e do t h i n g s i n t r i n s i c a l l y , i . e . on a f i n i t e dimensional

2 , -h

complex v e c t o r space, then t h e o p e r a t o r s

are

azj meaningless ( f o r they a r e defined i n terms of a b a s i s on azj

-

C"), but t h e d i f f e r e n t i a t i o n o p e r a t o r s

d,a,a

a r e meaningful;

df = af + % f r e s u l t s from t h e f a c t t h a t every C-

and then valued

l i n e a r form on a complex v e c t o r space i s , i n a

unique way, t h e sum of a complex l i n e a r form and a complex conjugate l i n e a r form, i . e .

y, : E +c,

if

x) = xp(x)

$1 then (#

cp(x-ty)

such t h a t

'p =

=

cp(x)

+ cp(y),

, for x

E R,

dJl + G2 $~(x+y)= ~ ( x )+ ~ ( y )

i s a complex conjugate l i n e a r form means

#(xx) =

X

for

Proposition 1. into

Co(U)

L E Q). The o p e r a t o r s

Proof.

#(U)

azj

from

1

C (U)

are l i n e a r , and s a t i s f y t h e product r u l e :

- -i h W(jf g )

= f

+(fg)

= f

j

Hence

and

az

azj

+ af azj g

%- + af azj

az

g, where

f , g E C1(U).

j

is an algebra of f u n c t i o n s .

Linearity i s clear.

Now, we have

d ( f g ) = fdg

+ gdf.


8

Hence

+

dz

-

1”

..,dzn,dz l,...,dzn

g ‘j=1

az

dz. j

+ af

j

azj

Since

d5.1.

J

a r e l i n e a r l y independent we g e t t h e

result. Proposition 2.

Every polynomial

I t i s c l e a r t h a t every constant function on

Proof.

i s holomorphic.

Since

l i n e a r function

an+

of

#(Cn),

Bn 3 6: i s holomorphic.

azh azj

C

=

0, f o r a l l

i s holomorphic.

h , j , then every Hence, t h e subalgebra

generated by t h e c o n s t a n t s and l i n e a r f u n c t i o n s , t h a t

i s t h e algebra of a l l polynomials on

an,

i s contained i n

Proposition 3 ( l o c a l c h a r a c t e r of a n a l y t i c i t y ) f 6 #(U),

V c U, open.

f : u+C, f E

Gn

U

=

IJ

j61

#(U). Proof.

Clear.

V

Then V jy j

flV

open,

#(V).

Let

Conversely, i f

f l V . g #(Vj), J

then

#(U).


93.

that

9

Cauchy I n t e g r a l . Let

f : U -76:

6

i s compact,

,U -

c C, open.

D c

Let

"Jordan domain of c l a s s Proposition 1.

C1

r a d i u s of t h e polydisc.

r = (rl,

boundary,

aoD c aD

0

,j

...,r n )

=

...,n ,

1,

a r e the

n

E R+ i s t h e poly-

r E R+, denote by

x...x

r

t h e polv-

-

Dr ( a n ) , [Er(a) = D (al)x n rl

t h e open, [ c l o s e d ] polydisc i n

=

J

D

1

6 , with

r;

1 '

D

,...,n .

= 1

Dr(a), [Dr(a)l t h e open, [ c l o s e d ] d i s c i n

a, radius

If

,j

D. c CC J

Dn c Cn

.. , r ) .

Dr(a) = D (a,)

radius

r.

r = (rl, If

x . . .x

i s a Cartesian product of

(Cn

If

(c.

r a d i i of t h e s e d i s c s , t h e n

center

is a

(n = 1). I f

D = D1

A Jordan polydomain

open (closed) d i s c s i n

Denote:

c o n s i s t s of

Sa D

An open (closed) polydisc i n

(r,.

(D

(Cauchy I n t e g r a l Theorem)

i s a Cartesian Product of Jordan domains

radius

aD,

D,

Cl").

D, then

D e f i n i t i o n 1.

-

D c U, and t h e boundary of

a f i n i t e number of Jordan Curves, of c l a s s

z

be open, such

an,

center

a = (al,.

...xDr- n (a,)]

.., a n ) ,

...,r n ) . XL...X~D n

a 0D, i s

i s a polydomain, i t s d i s t i n g u i s h e d

given by

(boundary of

a 0D

=

aD1

Y...x

D) , w i t h e q u a l i t y i f

aDn.

Note t h a t

n = 1.

Poly-


10

Proposition 2.

,U

f : U-C

(Cauchy i n t e g r a l , i n dimension

c Gn, open.

Suppose

n).

i s continuous on

f

p a r t i a l l y holomorphic w i t h r e s p e c t t o each v a r i a b l e . D1,..

. ,D n c Q:

D1 x . . .

z E D,

be open Jordan domains of c l a s s

U, and

Let

C1, and

O c U .

and

X D n = D ,

Let

(where i n t e g r a t i o n i s meant a s Cartesian product i n t e g r a -

tion). Proof.

Case

n = 1 --preceding p r o p o s i t i o n .

n = 2, f o r simplicity, ( a r b i t r a r y n = 1, twice.

t h e case when

n

L e t us take

i s similar).

W e apply

We have f ( z 1 Y Y2)

f ( z 1’z 2 ) = 2ni

dT2

Y2-z2

, since

a D2 (zl,

5) E

U

if

y2

a

D2, and

z2

D2

(since

holomorphic w i t h r e s p e c t t o t h e second v a r i a b l e ) .

if

y1

6

Dl

and

Y2 E

D2, and

z1 E D1.

Hence

f

is partially Moreover, w e


11

f ( Y19 Y2)

d

1

=

2

(2ai)

s aOD

f ( P1 9 Y 2 ) (y1-z1)(.y2-z2) dY1 d Y2

V1l

dy2 =

(by continuity).

HOLOI4ORPIlIC FUNCTIONS

12

54. Differentiation of Holomorphic Functions and the Cauchy Inequalities f : U ->a

Definition 1. Let on U, and a

(al,

=

If f 6 H ( U ) , then

...,a n)

-

=

and we denote this by

D

and

aaf -

If

la\ I. m y we set:

-

af af (since for analytic functions, az = az j j

Daf.

Proposition 1. Let

where

c Cn, open, of class Cm

a multi-index.

axa

aza

,U

f E #(U).

Then

f E Cm (U>, and

are as in Proposition 2 , 93.

z

Hence Da f E H(U). Proof. Assume n

2 fli

J

Y-2

dy

.

=

1. Then for z 6 D

, f(z)

=

By the classical rule of differentiation

aOD

under the integral, the integral in the right hand side is a Cm

function on

D. Hence, f is of class

COD

on

D. Moreover,


13

by the same rule, - -

dY

2ifi By the same rule,

2

D, and

aafa -

0,

ax

aaf

".

( a ) ( z > =

ax

D).

c >,function on

(in fact

is a c1

(z E

(Y-z)a+l

0 dY= 0. Therefore

-

E #(D),

axa aOD

a aaf ax

for all D. Therefore f E Cm (U), and The case when n Notation.

E #(U).

is arbitrary is analogous.

If z

Cn

,a

=

multi-index, then za

l a D

Remark 1. We can write then

f(z)

=

a!

where f(z)

a

+

. . . an+l).

1 = (al+l ,

s

= (-

fo

dY

.

For

=

1' z1

1 n (m) s

...znan .

(r-z)a+l

aoD

a = 0, we can write

(1 = (l,...,l)).

(Y-zIl

aOD

Proposition 2.

D

=

(Cauchy's Inequalities).

Let

f : U

Proof. Apply the preceding proposition, by taking 1 n Da f(z) = ( E ) D ( 2 ) . Then r a. (Y-2) r.=I Y . - z . I J J J j=ly.. ,n

4

s

.

letting

y. J

= z

+ r. ~

eiej, we get J

->

C,

, and

by

d 7

,


14

1 D' U!

f(z)

=

ie, 1

= (&

s

n

ie ,...,z

f(zl+rle

0 c

cm (u) c crn(v) c C ( U )

c

LT~~(u) =

(measurable functions on U which are locally bounded) c LToc(U)

=

(measurable functions on U which are locally p integrable) c 1 All the above spaces induce the same topology c Lloc(U) c D'(U). =

on I(U). Proposition 3 (Montel). compact (that is

x

A subset

x

c #(U)

is relatively

is compact) if and only if for all K c U,

compact, sup)f(z) I < m Z€K fEX

(that is, the functions in

X are


20

uniformly bounded on every compact subset of

U).

The Proof follows Lemma 6 , below. Lemma 1.

E Q (h,k = 1 , 2 , . . . ) such t h a t f o r a l l h,k h, t h e sequence k=1,2,... i s bounded. Then,

fixed

there e xi st s the sequence Proof.

a

Let

such t h a t f o r a l l f i x e d

k j = 1 , 2 . . J

.

Let

i s convergent.

...I.

= [1,2,

I.

Since

subset

I2 c I1

of

such t h a t

N

Lemma 2. that

-

t i o n s on

0

3

lirn k61h

Let

X = E , and

E

I

1 a

such t h a t

3

h,k

...

3

In

3

exists.

uk(k=1,2,

...)

Let

For every

x

W e thus

...

of i n f i n i t e s u b s e t s

Let

kl

c I1

X c E , such

x E X, then

E.

l i m u,(x) k

If exists,

E E , where we want t o prove t h a t

0

E

>

;

a sequence of complex func-

E , equicontinuous a t every p o i n t of

Proof.

lirn a k€Il 1,k

exists, etc.

be a t o p o l o g i c a l space,

l i r n uk(x) e x i s t s , f o r a l l k f o r a l l x EE.

exists.

lim a 2,k k€I2

such t h a t I

0

i s bounded,

CCn : l w j l

Izj

I
,

i s t h e extension

g

U we a r e looking for.

to

since

f o = (l-Q)f

g E Cm (U) ,

tu f i n d a "correction",

1 on a neighborhood

and i d e n t i c a l l y

g

be i n

qj E Cm (U),

We n o t i c e t h a t

ap.

Wi

i

j

&= aZ

and

it = 0

by l e t t i n g

we must have

g(U),

(j,i

=

outside

1,.

. ., n ) .

has compact Extend

@

j

U.

By Lemma 3 , we conclude t h a t t h e r e e x i s t s

g E COD (a?),

unique, w5th compact support , s a t i s f y i n g t h e above system; hence

U\K.

fo

-

g E #(U).

@.=

We have

Let us now prove that 0

[

on

i s holomorphic on we s e e t h a t

can s p t

continuation).

r flU #

Q,

-(j = 1

and s i n c e

Q,

g

,... , n ) .

on

Hence

g

has compact support,

vanishes i d . e n t i c a l l y i n t h e non-empty (open)

g

unbounded component

Hence

spt

an

J

f o - g = f

can s p t

of

C

d

r c

a(

Also,

fo

= f

Now,

a.

r

an

(O

(by uniqueness of a n a l y t i c

s p t )+ ,

on

cU

a

( s p t (p> c

spt

Q. Hence,

=

u. fo = f ,

DOMAINS

g = 0,

and

rn

on

r n U c U\K fo - g f

and on

5

U.

Hence,

u

If

c

f

empty; t h a t i s , u n l e s s

f

and

then

g

an

i s open,

-1

II n u.

on

f

n

E H~(u),

i s never compact

(0)

Since

2 2,

(unless i t i s

U).

never vanishes on

We s h a l l use the following f a c t from topology.

pm, such t h a t V

a r e open and connected i n

W

f

=

Q .E .D.

then t h e closed subset

V

-

(by uniqueness of a n a l y t i c c o n t i n u a t i o n ,

UVZ

Proposition 4 , ' $ 6 ) .

Proof.

fo

43

i s connected, then we conclude t h a t

U\K

Corollary 1.

OF HOLOMORPHY

If

U W =

wm,

n W i s connected ( s e e "Incidence Relations i n M i c o h e r e n t

V

Spaces", by A. H. Stone, Transactions Amer. Math. Society, 1949, Volume 6 5 ) . Nowj assume t h a t may assume t h a t

K

f

=

-1 (0)

i s non-empty and compact.

i s connected (otherwise , j u s t consider each

U

connected component).

Since

(of

a,

components being r e l a t i v e l y compact. r e l a t i v e l y compact components of d

with

i s open and connected.

U U K

u

V = U

, i s connected L e t now

N

K

and

(and cp

W @)

l/f

ccn g,

=

.

on

u n

0

by Hartogs' Theorem, But

(pf

t

1 on

U n

'p

kn

[0"

f

=

0

on

K

and

),

Let

C

K).

cn

K

Cn t h e remaining

$!

= K U (all

E

Then

i s compact,

Applying t h e above f a c t , we g e t t h a t

(U U

@

n

cn?

=

a:

2..

Since

cp E

#(u n

can be extended t o a f u n c t i o n

can 3, cp E

#(U).

N

K,

and so by uniqueness of a n a l y t i c

continuation (Proposition 4 , $ 6 ) , since

c

i s assumed compact,

K

has only one unbounded component

and

We

K # 4.

cpf

E

Q.E.D.

1 on

U,

a contradiction

u n

c Cn

DOMAINS OF HOLOMORPHY

44

$10

-

Domains of Holomorphy.

D e f i n i t i o n 1. An open subset

U 5 Cn

i s an open s e t of

holomorphy i f f i t i s impossible t o f i n d two connected , open subsets , U1 a)

U2

in

u2 #

qi.

and

u n u1

3

Cn,

with the following p r o p e r t i e s :

u1 d u. b)

For a l l

f E W(U),

unique) such t h a t Notice t h a t

aU

U n U1 # @,

and

f = fl fl U1

there e x i s t s on

# @,

f

1 E #(U1)

(necessarily

U2.

since

U1

i s open and connected ,

U1 d U.

I n t u i t i v e l y , t h e n o t i o n of an open s e t of holomorphy means t h a t t h e r e i s no non-empty p a r t of f E U(U)

aU beyond which every

may be a n a l y t i c a l l y continued.

Definition 2 . s e t of holomorphy

A domain o f holomorphy i s a connected open

.

C l e a r l y , an open s e t i s an open s e t of holomorphy i f and only i f each connected component i s a domain of holomorphy. Comment : Domains of holomorphy have important a p p l i c a t i o n s i n t h e o r e t i c a l physics.

See : A . S . Wightman , Analytic Functions

of Several Complex V a r i a b l e s , and V . S . Vladimirov, Methods of the Theory of Functions o f Several Complex Variables ( f u l l references a r e given i n the b i b l i o g r a p h y ) . For convenience, we introduce h e r e t h e following (redundant) definition.


Definition 3 .

U

C

45

open, i s s a i d t o be h o l o m o r p h i c a l k

Cn,

convex i f f o r every for a l l

~ ( u ) )is

f E

m: It

compact.

w i l l be seen i n Theorem 1, $11, t h a t

Proposition 1. Let for a l l

U'

3

q

E aU,

and

U

f'(rl) = 0,

U cCn,

7 E

and

U' ,

and

f'(z)

# 0,

open, connected, such t h a t

U' c C n ,

there e x i s t s a

open, connected, with

function for a l l

' c

f

#(U'),

Then

z E U.

where is a

U

.

domain of holomorphy Proof.

i s an open s e t of holomorphy.

U

holomorphically convex i f f

is

U

Assume t h a t t h e r e e x i s t s

U1

U2

and

satisfying

N

a ) and b ) of D e f i n i t i o n 1. of

n

U

connect Let

q

a

II

T,

containing and

b

U'

and

that i s

we move along fl

r, f'

E 3(U1),

1 f

7= fl

r

U2.

fl(z) * fl(?)

t h e r e does n o t e x i s t

U1

Hence

,,,

to

< and

m

then

, U2

Corollary 1. Every open s e t s e t of holomorphy

.

E

U2,

g

b,

to

z E

U).

1 f '=

fl

Let By b ) of on

f2.

But, i f

1 -m. f

a contradiction.

r

f = fl

on

(2)

U1.

where

7.

such t h a t

and

E U1\U

contained i n

for a l l

f l E #(U1)

on

from

from

g

corresponding t o

f(z) # 0,

(since

D e f i n i t i o n 1, t h e r e e x i s t s

U2;

r

by a polygonal l i n e

Choose

E H(U)

be the connected component

Take a point

U2.

be the f i r s t point of

dU.

meets f =

U1

U2

Let

But s i n c e Therefore,

s a t i s f y i n g D e f i n i t i o n 1. Q.E.D.

U cC

(n = 1 )

i s an open


46

Proof.

U'

=

and

(c,

U

Assume t h a t

=

f'(z)

z

-

i s connected.

for a l l

q,

1 E aU,

Notice, i f

then we may

z F C,

apply P r o p o s i t i o n 1. This C o r o l l a r y e x p l a i n s why domains o f holomorphy

Comment.

are n o t of i n t e r e s t f o r one complex v a r i a b l e . Definition 4 .

Let

U cV

We say t h a t

a n d connected.

i f and o n l y i f every

I n o t h e r words, t h e mapping

Proposition 2 .

If

g F w(V)

~ n t o~

#(V)

algebra i s o m o r p h i s m Y i ) .

We say t h a t

U c

every a n a l y t i c c o n t i n u a t i o n

U

V

-

glU E #(U)

( u )(it

is

i s always a n

i s proper i f

V # U.

i s a domain o f holomorphy,

(Cn

V

a r e open

V

has a ( n e c e s s a r i l y unique)

g E #(v).

an a l g e b r a isomorphism from

and

i s an a n a ly tic continuation of

V

f E #(U)

analytic continuation to

U

where

c (c",

of

U must be

=

U

(i.e.

U

has no proper a n a l y t i c c o n t i n u a t i o n ) . Proof. V.

rake

Ul

of holomorphy

Assume t h a t

.

Example 1. q E V.

V

Then

U2

= V,

U

U has a proper a n a l y t i c c o n t i n u a t i o n t o conclude t h a t

= U

U

i s n o t a domain

Q.E.D.

Let = V

n

2,

- {,,I

V c Qn open and connected.

i s n o t a domain of holomorphy, because

i s a proper a n a l y t i c c o n t i n u a t i o n of

U

(by H a r t o g s '

Theorem, $9 , Theorem 1). I n a more g e n e r a l way, l e t open and connected, and l e t V\K

since

i s connected. V

Then

Let

U

K cV = V\K

be compact,

# a,

V c En

be

such t h a t

i s n o t a domain o f holomorphy,

i s a proper a n a l y t i c c o n t i n u a t i o n of

U.

DOMAINS OF HOLONORPHY

47

The converse of Proposition 2 i s not t r u e ;

Remark 1.

t h a t i s , t h e connected open s e t

U

may have no proper a n a l y t i c

c o n t i n u a t i o n , withbut being a domain of holomorphy.

In t h i s c a s e ,

the envelope of holomorphy

an

e(U)

f a i l s to exist i n

(See

Remark 3 , g 1 3 ) . Proposition 3 .

U

Every convex open s u b s e t

of

is a

(En

.

domain of holomorphy

We s h a l l use t h e following (Hahn-Banach)

Proof.

s e p a r a t i o n theorem on a f i n i t e dimensional r e a l v e c t o r space if

U

C

i s open, convex, and i f

E

a r e a l - l i n e a r form x E

on

E

7 E E,

k

U,

we g e t a r e a l - l i n e a r form

> v(z),

Q(z)

= p(z)

-

for a l l

z E

i(,,(iz),

for a l l

i s complex-linear.

rc)

because by taking

Re $ (7) =

U' = E n ,

t o conclude t h a t Remark 2 . take

for a l l

u.

p(?)

so

then t h e r e i s

such t h a t cp(,,) > ,(x),

an.

Apply t h i s t o the r e a l v e c t o r space q

, k U,

E:

n = 1,

U

U.

: C n +IR,

z E

Moreover,

v ( ~ )>

-

an.

Then

by

rc)(iz) = i $ ( z ) ,

# rc)(z),

rc)(,)

E Cn,

such t h a t

n rC, :(I @

Define

,

So, i f

for a l l

z E U,

Re $ ( z ) = c p ( z ) . Apply Proposition 1,

f'(z) = $(z)

-

for a l l

rc)(n),

z E

an,

i s a domain of holomorphy.

The above proposition does n o t i n c l u d e , i f we t h e f a c t t h a t every open s u b s e t of

open s e t of holomorphy. Definition 5 .

CC

i s an

This suggests t h e following e x t e n s i o n :

An open s u b s e t

U c

an

i s subconvex i f f o r


48

all

U,

q

an,

E

t h e r e e x i s t s an a f f i n e complex sub-

manifold

S c Cn

of codimension 1, containing

from

(Note:

By an a f f i n e complex submanifold i s meant

U.

Cn).

a t r a n s l a t e of a complex v e c t o r subspace of Every convex open subset i s subconvex. of

C

i s subconvex

(take

S

= {q]).

Every open subset where

C\S,

i s an a f f i n e complex manifold of dimension

and d i s j o i n t

5n -

S c Bn

2,

i s not

subconvex. Proposition 4 .

U

Every subconvex open subset

of

Cn

i s an open s e t of holomorphy. Same a s f o r Proposition 3 .

Proof.

Remark 3 .

This p r o p o s i t i o n ,

every open subset of

for a l l

,

X E A

shows t h a t

i s an open s e t of holomorphy,

Ux c Cn

If (

n = 1,

i s an open s e t of holomorphy.

CC

Proposition 5 .

for

some indexing s e t ) ,

and

U

i n t e r i o r ( fl Ux),

=

LEA

then

U

i s an open s e t of holornorphy.

Proof. respect t o

U.

u1 C ux.

Since

f

x

on

and

U2

U1 & U,

Since

b e a s i n D e f i n i t i o n 1, with there e x i s t s

Let

f o r which holomorphy

f = f

f = fl

.

x

E I\- such t h a t

Ux i s an open s e t of holomorphy, t h e r e e x i s t s

such t h a t t h e r e i s no

E #(U,)

U2.

U1

Let

I

x on

Q .E .D

.

U E #(U). U2.

f l E #(U1)

f o r which

Then t h e r e i s no

Therefore

U

f

f

x

= fl

E #(U1)

1 i s an open set of

49


311

-

The Cartan-Thullen Theorem Definition 1. If X c U can, U

hull of X with respect to f; U

=

Ez : z E

u,

open, we define the A

denoted by

#(U),

Xu,

as

5 sup

If(z)l

WEX

It is immediate that: 1) X c 2)

3)

fi,

f,

cU

is closed in U A

A

= $,

!bu

uu

if X1 c X2 c U

5)

if X c U c V su

If(x)l

is the largest set Y c U,

for all f E

A

then X1 (U,V

for which

,

^ %

then Xu c

for all f E #(U>

suplfl

=

suplfl ,

Y

~(u) 8)

Remark 1. If f E #(U), If(z)l

I c y for all z E

If(z)l

c, for z E X ,

9,.

xu A

=

h

(c,

for all z E X, then

That is, any estimate of the form

extends to the same estimate on

2

X

(closure in U)

Xu

If(z)l

Proposition 1. If X c an, the closed convex hull

U open)

sup If(x)l

=

A

c X2u

x EX

X&

2,

u

4)

6)

In fact

=

4J .

A

then XCn is contained in

of X .

Proof. Let us use the following (Hahn-Banach) Separation Theorem q

real finite-dimensionalvector space E: If X c E,

E E, then

q E

the closed convex hull of X

real-linear forms cp on E, we have

iff for all

c p ( ~ ) 5 sup cp(x).

x FX


50 A

q = (TQ,...,,),

E Xcn,

Now, take

where c z

-

= a ib and d e f i n e f ( z ) = e j j jy n z = ( z l y . . . , z n ) E CC Then f E 3(Gn), and

Let

c

.

TI.=

+...sc -2J

n,

[.J + i p j . for a l l

5 sup

If(q)l

If(z)l;

z EX

that i s , e that is ,

+... anxn sbny n

al%*lh+****nL*npn

-

.. h n % s b n p n

alS1Sblp1+.

a x sb y sup e 1 1 1 1

~

5

ZaC

+. . .+anxn+bnyn ) .

sup(a x sb y 11 1 1 X

Therefore, by the Separation Theorem, closed convex h u l l of

i.e.,

X,

(

cl,. . . , E , , p l , . .. , h ) E

A

Q.E.D.

EX.

Note:

cp

We have used here the f a c t t h a t any r e a l - l i n e a r form m on Rm i s of the form c p ( x l , . . .xm) = c aixi, where ai E IR. i=l Corollary 1. I f X c Cn i s bounded then Xcn i s compact. ~

More g e n e r a l l y , i f

9,

bounded, then Proof.

If

follows using Remark 2 .

ft,

because since

$,

then

A

n

Xu c X ,

and i f

A

h

xu

2

i s bounded then

X

9,

even i f

The r e s u l t

Property 2 and Proposition 1.

c icn c X,

Although

i s bounded.

X

i s closed i n

U, i t may f a i l t o

i s compact and contained i n

may f a i l t o be compact (hence t o b e c l o s e d i n

,

i s bounded).

U = C n - Eq],

and

X

=

compact).

Then, i f

extension

f E #(a?),

5

f E #(U),

Cn,

-

we know t h a t

f

has an a n a l y t i c

so t h a t by t h e maximum modulus theorem, n

suplf(,

x

U,

For i n s t a n c e , l e t E On (n 2 2 ) , n [ z F Cn : c I zi qiI2 = 1 3 (X i s

i=1

we have I f ( z ) l

X is

i s bounded.

Cn,

be closed i n

U open,

X c U,

for a l l

z

such t h a t

C Izj

j =1

-

qjlL

5

1.

.Y


n

Thus, every

q,

Xu

n o t compact

If

-

eU.

I 1,

(since

f(z) E C

t h e semi-norm

can,

K c U

i s t h e s e t of a l l

f E #(U)

2

qjl

belongs t o so 9, i s

6! U),

.

Proposition 2 .

&

-

c Izj

with

j =1 i s n o t closed i n an

A

Hence

#

z

51

z E

on

f E #(U)

compact,

U

open, then

such t h a t t h e l i n e a r form

U

i s continuous, with r e s p e c t t o

x(U) +

K

sup If1 Elk,.

K The proof of Proposition 2 i s based on t h e following l e m a

cn normed a l g e b r a s : Lemma 1.

Let

A

be an a l g e b r a , endowed with an a l g e b r a

semi-norm (a v e c t o r space semi-norm, with t h e a d d i t i o n a l property I(xylI cp

5

lIxII.llyll).

If

c p : A -6:

i s continuous i f and only i f

i s a homomorphism, then

I IYl I 5

Proof of Lemma 1. C : C l e a r l y cp

i s continuous.

c

3 : If

Iydx)l

p i s continuous,

.

implies t h a t then t h e r e e x i s t s

= I,(xn>l

1 Ix

Ic

i s an a l g e b r a homomorphism, and norm.

f

-

sup1 f

I

i s an algebra semi-

K

Therefore, Lemma 1 a p p l i e s . Remark 3 .

on

I IYl I 5 1

0 such t h a t

IydX>l"

all

1.

z E

K,

The proposition says t h a t

U

such t h a t i f

then

f ( z ) -, 0 .

f E #(U)

A

%

i s t h e s e t of

v a r i e s , tending uniformly t o 0


52

Definition 2 . 7

i s a r e g u l a r point f o r

such t h a t

fl E 3(U1),

n e c e s s a r i l y unique).

Ul, U2

exist.

f

U n Ul

and connected, with and

f E u(U)

Let

au

U2 # 6 ,

f

fy

=

f

n

f l 6 #(U1)

with

f E

U1

~ ( u ) which

3

f

q E U1

a r e singular f o r

U = @ a n d Ul 2

1

U2.

on

= f

(Clearly,

U),

fl

is

i f no such

aU

(Cn,

f

open, connected,

i t i s impossible t o f i n d

U,

q!

Denote

by

Su # 0

By a l a t e r r e s u l t (see $13),

bu

f

k

U1

In o t h e r words,

U1,U2 c

U2.

on

f.

a r e everywhere s i n g u l a r on

o f holomorphy i f f

(hence

open

i s everywhere s i n g u l a r on

i s everywhere s i n g u l a r i f f o r any such t h a t U

Ul,U2 c Cn,

i f there e x i s t 3

We say t h a t

i s a singular point f o r

7

We say t h a t

i f a l l points of

7 E aU.

and

Su

the s e t of a l l

au. iff

i s an open s e t

U

w(u>.

i s of f i r s t category i n

#(U) Theorem 1.

U c Cn,

For a connected open subset

t h e following conditions a r e equivalent :

1) U 2)

(Cartan-Thullen).


i s compact, then d i s t ( K , CU) = d i s t ( $ ,

K c U

If

r e l a t i v e t o t h e norm

I IzI I

=

sup Izj

I,

where

CU),

z = (zl,.. .zn).

j 3)

U

i s holomorphically convex; t h a t i s , i f

compact , then

$f

i s compact.

4 ) There e x i s t s s i n g u l a r on

au;

Remark 4 .

Su #

such t h a t

f

i s everywhere

@.

In condition 2 , we use a s p e c i a l norm on

( j u s t because of the use of polydiscs). remains t r u e f o r any norm on

is

(Compare with Remark 2 . )

f E #(U)

that is,

K c U

(En,

8

Actually, c o n d i t i o n 2

even i n the following form:

if


v

i s t h e l a r g e s t balanced open s u b s e t of

(K

+ V) n

[U

we a l s o must have

= @,

Remark 5 .

53

an

such t h a t

($ + v > n

$lo),

c o n d i t i o n (b) s a y s t h a t , f o r a l l

a s d e s c r i b e d i n t h a t d e f i n i t i o n , t h e r e i s some

U1

with

and

U2)

f = fl

U2.

on

such t h a t f o r a l l

U1,

f E w(U),

U2

(depending

f l E #(U1),

u1

and

as d e s c r i b e d i n t h e d e f i n i t i o n o f domain

fl E #(U1)

with

f E u(U)

be a n a l y t i c a l l y continued along any p o i n t o f

U, b u t n o t

H(U)

(not depending on any

In o t h e r words, t h e r e e x i s t s some

Remark 6 .

(

Ul, U2

On t h e o t h e r hand, c o n d i t i o n 4 i n Theorem 1

o f holomorphy, i t i s impossible t o f i n d

U2.

f

such t h a t i t i s impossible t o f i n d

s a y s t h a t t h e r e i s some

on

= @.

In t h e d e f i n i t i o n of domain o f holomorphy

(Definition 2 ,

on

Cu

f

=

which cannot

aU.

I n Theorem 1, c o n d i t i o n 3 i s t h e o n l y one u s i n g

c..

Condition 3 i s i m p o r t a n t , t h e r e f o r e , because

i t allows t h e d e f i n i t i o n o f S t e i n manifolds.

Open s e t s i n

mn

a r e g e n e r a l i z e d by d i f f e r e n t i a b l e m a n i f o l d s .

Open s e t s i n

Cn

a r e g e n e r a l i z e d by complex a n a l y t i c m a n i f o l d s . holomorphy i n

Cn

Open s e t s o f

are g e n e r a l i z e d by S t e i n m a n i f o l d s .

To prove t h e Cartan-Thullen Theorem, we need : Lemma 2 . Dr(0) that q

Let

K c

f E N(u),

r

a n open polydisc w i t h K

+ Dr(0)

c U.

If

converges normally on

r)

E

Dr(q)

j

u can,

> 0,

G, =

for

K compact,

u

open,

j = l,...,n,

t h e Taylor s e r i e s o f

+ Dr(0).

fl

such f

(Recall h €A

at

fx

U2),


54


c sup L€A K

IfJ

O

1 1 ~ I1 = U

€>O in

= { z ~ U : d i s t ( z , C U )>

Clearly,

0 E U

and t h e r e f o r e E' U may be joined t o

Since i n some

U

P

U

u

=

and

U

E

0

be t h e connected component of

E

z

€

We may assume t h a t

j

U (0) U =

suplzjl.

U

in

UE(0):

=

El zll],

U

€'

Let

UR(0). where Let

We have

I n f a c t , every p o i n t

€>O 0

by a polygonal

and hence i n some

(We used h e r e t h e f a c t t h a t

U

UE i s open.

i s compact, connected, and c o n t a i n s u€

0

U

Uc(0), I>

€1

u

P

contained i n

0,

and hence

c2). €2

P

U.

i s contained z E some U ( 0 ) . E


64

whe r e

TE

= { t : It.

J

I
1, and s o

=

c

f = g

t h e n , as b e f o r e

fu(z)

0,

c l o s e enough t o

(tlz l , . . . , t z ) E U n n

i s close

z

g(z) = f(z).

UE(0)

on

1 . 4tn

norma 1l y

’

un+l )

*

1

fu(z) =

2 n1

i s holomorphic on we s e e t h a t

modified Cauchy i n t e g r a l .

UE(0).

f (z) = U

Hence, as b e f o r e

Hence p u t t i n g e v e r y t h i n g t o g e t h e r ,

-,

Moreover, f o r by t h e

a!

fJz)

=

D‘f(0)zu

Remark 3 .

hence on

U,

as

E

+

0.

Q.E.D.

This i s t h e e x t e n s i o n t o s e v e r a l v a r i a b l e s o f

t h e following theorem f o r one complex v a r i a b l e : U

open, and l e t

f E #(U).

Let

U.

0 E U c CC,

Then t h e Taylor s e r i e s f o r

converges normally on t h e l a r g e s t open d i s c , c e n t e r contained i n

on

4 D‘f(0)zu,

f(z) =

a20 UE(0),

since

i s a Reinhardt

U

1

Ul+1

f(z)=g(z)=

On t h e o t h e r hand, i f we s e t

normally on

I , and

IzI

Also, i f

g E H(UE(0)).

(tl

UE(0).

€1

t h e modified Cauchy i n t e g r a l shows t h a t

UE(0) Now ,

z

=

By the u s u a l d i f f e r e n t i a t i o n under

E a,TE.

t h e i n t e g r a l , we s e e t h a t

for

+ E)z)

( i . e . a connected Reinhardt s e t ) , so t h a t

t = (tl,..

Since

Note t h a t t h e

0,

f


Definition 6 .

Let

65

be a Reinhardt s e t .

X cBn

n r e p r e s e n t a t i o n i n R+, denoted

1x1

is

Its

real

rlzl = ( ~ z 1 ~ , . . . , ~ z: nz ~E) X I .

C l e a r l y , t h e r e i s a n a t u r a l one-to-one correspondence between Reinhardt s e t s i n Example 1.

and subsets of

(Cn

1x1

that

A connected open s e t which i s n o t a

(Hartogs).

domain of holomorphy:

Let

a holomorphy domain. Taylor s e r i e s

lzll

c za

a

0 . Consider the f i n i t e union of j the corresponding closed polydiscs X = U Dr(0), and l e t r €R with a l l

r

V

V

., .znn

f ( z ) = czl 1

such t h a t


66

If(rl)l > 12 suplfl

i f and only i f

X

?

l e a s t log-convex Reinhardt set Note:

%

It i s c l e a r t h a t

q

of

does not belong t o the cCn

containing

X.

e x i s t s , s i n c e an a r b i t r a r y

i n t e r s e c t i o n of log-convex Reinhardt s e t s i s a log-convex Reinhardt s e t . Proof .+:

If

f

e x i s t s , notice t h a t

{z E

(cn

: If(z)l

17 4

i s a log-convex Reinhardt set containing but n o t containing :

# 0.

all

Hence

7.

q

hence containing

X,

X,

4 %.

Conversely, assume

v d?.

Assume f i r s t t h a t

Consider the f i n i t e s e t

n log R = {log r = (log r l , . . . y l o g rn) E R : r E R J .

r

Let

n '?

+ at:]

(log1 (j =

be the convex h u l l of log R .

1 ...,n),

and so

,-.

E

=

my

modularly decreasing

Y

=

(yl,

ey E (X

f is

compact.

We have

15

f o r otherwise we would have log1 nj

f o r some

5, s i n c e

Then

...,yn)

F

r.

~ h u s IqjI

x"

(by log-convexity) and

5

yj

e 'j ;

is

being modularly decreasing) , a c o n t r a d i c t i o n .

By t h e Hahn-Banach s e p a r a t i o n theorem, t h e r e i s a l i n e a r form cp

on Rny with p o s i t i v e c o e f f i c i e n t s , such t h a t

( I f we have c l o s e d , with such t h a t

convex and compact ,

K c Rn K

n

F = p Y then t h e r e i s a l i n e a r form F = $.

cp

on Rn

+,R;I

then

K

c p - must have p o s i t i v e c o e f f i c i e n t s and

By d e n s i t y of

r

F c Itn, convex and

i n f .I > sup cp. It follows t h a t i f

F

p(logl,,l)>sup cp.

Q,

in

R+,

positive rational coefficients.

( ~ ( 1>) sup 9). K

we may assume t h a t

has

By m u l t i p l i c a t i o n by a l a r g e


has p o s i t i v e i n t e g e r c o e f f i c i e n t s .

i n t e g e r , we may assume

That i s , t h e r e a r e i n t e g e r s

X

E

= SUP

+... vn tn '

f o r every

c = e - X , we see t h a t

cn

satisfying

Next, i f

< n,

,.. . ,vn

0

and

lzll

1 z E X).

... --

preceding c a s e t o

,...,rn) -

5 r ,.. . ,lznl 5 rn, f o r some r Then take

all

E R

. . . z nn .

v1 f ( z ) = czl

has some of i t s coordinates = 0 , say e x a c t l y

q

q1 -

say

E

...nzI >

2

X

r . Setting 1 2 Icz:l ...zvnl , for n

t = ( tl , . . . , t n )

1cG1

(in particular, for a l l

k

v1,v2

R, such that v1 10glqll + . . . + v n 10gl')nl >

v 1t 1

z E

61

qk

(qk+l,.

= 0,

.

q k + l y . . ,tl, # 0, j u s t apply the

. . ,%) ,

to the f i n i t e s e t of a l l

r = (rl,...,rk,rk+l,...,rn)

(where

E R),

and

(k ' +In

to

cc

and

Q.E.D.

IRn-k.

Proposition 4 .

Let

U cCn

be open.

Then the following

conditions a r e equivalent:

1) U series

c c z a in a

2)

i s the open s e t of convergence of some convergent

U

c".

a

i s an open connected Reinhardt s e t containing 0 ,

and a domain of holomorphy.

3)

U

i s open, contains 0 ,

i s modularly decreasing and log-

convex. Proof.

1+ 3 :

The only thing t o be v e r i f i e d i s t h a t the

log-convex.

c c za is a a To t h a t end, we show f i r s t t h a t the s e t of

boundedness

B

open s e t of convergence of a convergent s e r i e s

i s log-convex; t h a t i s

IBI

i s log-convex.


68

So, l e t

x

r

> 0,

I.(

=

> 0 and

2 0.

s

and

X+p = 1.

...rnan 5 C,

(calrl

=

...snch 5 C ,

al lcalsl

X,

Raising the f i r s t t o the power

power

and m u l t i p l y i n g ,

p,

for a l l

0.

a

Hence

E IBI

(sl,...,sn)

,

Therefore, t h e r e e x i s t s

n.1

such t h a t a

( rl , . . . , r n )

we g e t

for a l l

t h e second t o t h e

lcaI (r;sf)'l..

r x s p E IBI.

> 0

C

.(rxs'l)a" n n

B y and hence

Thus,

C,

6,

i s log-convex.

3

+ 2:

Since every modularly decreasing s e t i s a

connected Reinhardt s e t , a l l we have t o v e r i f y i s t h a t

U

a domain of holomorphy.

is

holomorphically convex ( s e e $10, D e f i n i t i o n 3 ) . There e x i s t s a f i n i t e s e t

be compact.

U

To t h a t end, we prove t h a t

R

is

K c U

Let

r

of p o i n t s

=

( rl,...Fn)

L__

with a l l

r. > 0 J

such t h a t

s u f f i c i e n t t o prove t h a t

U.

Now

To show t h a t

C X n, d

U.

So,

be compact, as follows:

Let

convex combinations of then

Q

+,(I A

?u

by log-convexity of

1 k Er , . . . , r 3,

Dr(0) c U

U

(since

U

r ER

i s modularly d e c r e a s i n g ) .

closed i n

K C X =

i s compact, i t i s

i s compact, s i n c e

$,

is

J . .

6

an

A

cX d

(by Lemma l), Now,

Xu c X .

Y

ft

ly

X

and

i s e a s i l y seen t o

be t h e s e t of a l l m u l t i p l i c a t i v e

p o i n t s of

(That i s , i f

R.

R =

Y =

Then

Y

i s compact, as t h e continuous image of a compact s e t .

Then

ii'

i s seen t o be t h e s e t of a l l

Iz

1 5 1 TI,

f o r some

q

Y.

cU

Hence

z E Cn d

X

such t h a t

i s compact.

Since


2U

x" c U

c

fir

2U

and

i s closed i n

69

A

U,

i s compact. Therefore

Xu

i s compact. 2 3 1): Since

i s a domain o f holomorphy, by c o n d i t i o n 4

U

of the Cartan-Thullen Theorem, t h e r e i s a f u n c t i o n

f E w(U)

which cannot be a n a l y t i c a l l y continued beyond any point

0 E U,

Since

c c za

take the Taylor s e r i e s

a

We know t h a t

C c za a

f

U,

since

0.

at

converges (normally) on

U

a

Now, the open s e t of convergence

c c

of

V

U

U.

Hence

V = U,

dU beyond which

a point of

of

a

i s a connected, Reinhardt domain containing

and contains

E dU.

f

(by Proposition 3 ) .

0

i s connected

za a

f o r otherwise

V

would c o n t a i n

could be a n a l y t i c a l l y continued.

Q.E.D. Remark 4 :

In the above proof

, we

used

2

d

an

c X.

This i s ,

i n f a c t , an e q u a l i t y . Remark 5 .

3)

+ 1)

We used here t h e Cartan-Thullen Theorem.

However

can be proved d i r e c t l y , without using t h e Cartan-

Thullen Theorem : Let

R

be a f i n i t e s e t of points

r

=

E U,

( r ,...,rn)

1 r4 > 0,

with a l l

and s e t

J

q

C

4

2,

for

...z nn

v1 Z

~

deg f > 0 . by replacing near 1.

nl

X c U.

X =

U D _ ( O ) c U. If rER By Lemma 1, t h e r e i s a monomial

d U,

then

f(z) =

.

I f ( q ) l > 12 supif We must have X W e may a c t u a l l y assume t h a t f(qN > 1 > SUPlfl, such t h a t

f

by

Bf,

where

0

. J j For each f consider the corresponding f E #(U1) (for j y jl on u2>. BY assumption, I f j l l ( m on ul, which f j l = f j . By Montel’s Theorem ($5, Proposition 3 ) , we for j = 1 , 2 ,

So, l e t

-+

...

can e x t r a c t a subsequence from

u1

every compact subset of

L m

U1.

converging uniformly on f l E g(U1).

t o some

Clearly

on U2’

f. = f on U2’ we g e t f = f l J jl by passing to the subsequence. Hence f F %(U,U1,U2),

and so

I ~ , ( U , U ~ , U ~ )i s closed.

lfll

on

From

We next remark t h a t

i s dense i n

[s(UyU1,U2)

#(U);

t h a t i s , dense i n

~ ( U , U l , U 2 ) . Assuming t h i s , we conclude

that

i s nowhere dense.

w,(U,U1,U2)

i s dense i n set

g(U),

U

cause

dense i n

#(U,U1,U2)

But this follows from the

a(U).

i s a proper v e c t o r subspace of

g(U)

(be-

i s an open s e t of holomorphy) and t h a t on every topolo-

g i c a l v e c t o r space subspace

F

F c L F .

Let

E

~w,(UyU1,U2)

i t i s s u f f i c i e n t t o show t h a t t h e smaller

[ H ( U , U ~ , U ~ i) s

fact that

To prove t h a t

E,

the complement

i s always dense. b € F andlet

CEF

of a proper v e c t o r

(Cn f a c t , i t s u f f i c e s t o show a € C F .

E

Not i c e t h a t


73

1 1 b + T a - b a s j -‘OD and b + T a t C F ) . J J E A Next, we remark t h a t L S = u #(U,U1,U2).

On t h e o t h e r

U

J2 f

hand, i f open b a l l

belongs t o some

then t h e r e e x i s t s an

Qn whose c e n t e r has r a t i o n a l c o o r d i n a t e s ,

in

V1

#(U,U1,U2),

whose polyradius has r a t i o n a l c o o r d i n a t e s , whose c e n t e r l i e s i n N

of

i s compact and contained i n

U1, b u t such t h a t

U2

in

U n U1,

the connected component: U2

t h e connected component of the c e n t e r of let

0

m

be an i n t e g e r g r e a t e r than

V1

in

sgp If

I

and whose c l o s u r e V1

d U.

U n V1 (
1 > supif, J

J

L

J

I.

L3 =

c U,

i s contained i n

tj

L and z1,z2,. E X such t h a t z 4 and j y j U if j < k . Choose f . E H(U) such t h a t I f . ( 2 . ) I > sup J J J L. By multiplying by a s u i t a b l e s c a l a r , we may assume some

U.

=

A

U

%

Let

(U b e h g holomorphically

there exists X

Hence

U.

be c l o s e d , b u t n o t compact.

U convex)

Hence, every

be compact s u b s e t s such t h a t every

cU

U

compact subset of

$f

f E w(U)

i s holomorphically convex.

X CU

Let

. . . c Kn L1

U

Every

I < OD .

i s bounded on the closed s u b s e t

f E a(U)

Define

compact.

sup If I = sup If

s a t i s f i e s the c o n d i t i o n

K1 c

K c U,

z

j

‘Lk

Ifj I-

BY r a i s i n g t o a s u f f i c i e n t l y l a r g e

i n t e g e r power, wej may assume, i n d u c t i v e l y , t h a t

suplf. J

I

1 2j


f (z.)l k J

-

If(zj)l

as

a,

Comment.

j

If

4

a,

,

and

75

1 = 2k+l

5 c

i s unbounded on

f

Hence

2j

k>j

Q.E.D.

X.

i s any completely r e g u l a r t o p o l o g i c a l

U

space, then i t i s known t h a t t h e following a r e e q u i v a l e n t :

1)

c(U)

2)

A closed subset

every

f

c(U)

(= space of a l l c o n t h u o u s f u n c t i o n s ) i s b a r r e l e d .

i s compact i f (and always only i f )

X cU

i s bounded on

Lemma 1. L e t

U

X.

be open, connected, and non-empty.

Cn

Assume t h a t t h e r e e x i s t s an open connected s e t

U

an a n a l y t i c c o n t i n u a t i o n of

i s contained i n

U

V,

Then

and hence

i s unique. Proof.

Assume t h a t

continuation of

U # g, U

h i c h is

and a domain of holomorphy.

every a n a l y t i c continuation of V

V c Cn

V

fl

W

3

U,

~(w), we

a domain of holomorphy.

U

and

Let W

W

Hence,

be an a n a l y t i c

a r e open and connected,

a(V)

and every function i n

with a f u n c t i o n i n

continuation of

U

Since

U.

exists.

V

must have

c o i n c i d e s on

W c V,

since

V

is

i s the l a r g e s t a n a l y t i c

V

and, as such, i s unique.

By Lemma 1, t h e following d e f i n i t i o n i s meaningful. D e f i n i t i o n 1.

Let

U c Cn

The envelope of holomorphy a n a l y t i c c o n t i n u a t i o n of

e(U) U

in

be open, connected, and non-empty. of (Cn

U

in

(Cn

i s t h e unique

which i s a domain of holomorphy,

i n case such an a n a l y t i c c o n t i n u a t i o n e x i s t s .


76

Remark 3 .

Cartan and Thullen gave an example of an open,

U

connected s e t

(c

2

,

U # 9,

in

EL.

U,

may f a i l t o e x i s t i n

The reason

e(U),

f o r which

e(U)

does n o t e x i s t

f o r some open connected non-empty set

an

i s because when we t r y t o enlarge

V

U by connected open a n a l y t i c c o n t i n u a t i o n s

of

U.

in all

p o s s i b l e ways , we may end up with multiple-valued f u n c t i o n s . (See Vladimirov, Methods of the Theory of J?unctions of Several Complex Variables, page 1 7 7 ) .

This i s due t o the f a c t t h a t we

forced o u r s e l v e s t o consider only single-valued holomorphic

an.

functions and open, connected subsets of

To b u i l d a more

s a t i s f a c t o r y theory, we must then give up considering only open, connected subsets of

En

and pass t o complex, a n a l y t i c manifolds,

and among them, Stein manifolds.

The idea of enlarging

U

in

a l l possible ways leads u s t o the concept of a Riemann domain, t h a t i s , a connected topological space

R

together with a l o c a l

homeomorphism of i t onto an open, connected subset of Remark 4 .

Given a domain (open, connected,

consider the connected component

E(U)

of

U

# 9)

(notice that

U,

so

E(U)

Cn

always e x i s t s ) .

U c E ( U ) c e(U)

containing

an,

En

containing

i s i t s e l f a domain of holomorphy containing

of holomorphy containing then

U c

i n the i n t e r i o r

of the i n t e r s e c t i o n of a l l domains of holomorphy i n

U

Cn.

U) , and so

Then

U. I n case

(since E(U)

e(U)

E(U) e(U)

i s the s m a l l e s t domain does e x i s t i n

(Cn

i s a domain of holomorphy

i s an a n a l y t i c c o n t i n u a t i o n of

U


By Lemma 1, t h e r e f o r e ,

which i s a domain of holomorphy.

-

e(U)

If

E(U).

e(U)

an,

does n o t e x i s t i n

i s never an

E(U)

unfortunately

77

then

a n a l y t i c c o n t i n u a t i o n of

U,

and so i s of l i t t l e i n t e r e s t . Proposition 3 .

0.

containing log-convex

Then

e(U)

Since

exists i n

W

V.

Then


V

is a

U.

By Proposition

We claim t h a t

U; hence

e(U)

W

is

e x i s t s and equals

c c za

Take i t s Taylor s e r i e s

f E #(U).

containing

i s a domain and i t i s t h e l e a s t log-

W

an a n a l y t i c continuation of Let

V

be t h e l e a s t modularly-decreasing set

W

kt

convex , modularly-decreasing s e t containing

4 , $12,

U.

s e t containing

i s a Reinhardt domain, we s e e t h a t

Reinhardt domain. containing

and i t i s t h e l e a s t

Cn

Consider the l e a s t log-convex s e t

U

an

be a Reinhardt domain i n

, modularly-decreasing

Proof.

U.

U

Let

at

0.

W.

Since

a C L C c

a

za a

converges t o

we s e e t h a t

U

i s contained i n t h e s e t

c c za. Hence W a

c By

a

log-convex.

a t every point of

f

So,

since

t;

W c

B

(because i t does on

c

that is

continuation of

f

continuation of

U.

to

B). W

Q.E.D.

(by Proposition 3 , § 1 2 ) ,

of boundedness of

i s modularly-decreasing and

a

W

B

U

c za

a


Thus t h e s e r i e s d e f i n e s an a n a l y t i c from

U.

Hence

W

i s an a n a l y t i c

78

LOCAL PROPERTIES

Chapter 3 :

$14.

Local P r o p e r t i e s

Germs o f A n a l y t i c F u n c t i o n s . D e f i n i t i o n 1.

Let

t h a t a f u n c t i o n i s a n a l y t i c ( o r holomorphic)

on

a n a l y t i c (holomorphic) on some open s u b s e t o f

wx

Den0 t e by

U c

Notice t h a t

%

f :U

the s e t o f a l l functions

on an open s e t

an

U

such t h a t

i s fixed, but

X

8.

be any s u b s e t o f

X c En

3

We say

i f it is

X

an

containing

C

holomorphic

+

X.

X.

U v a r i e s with

f;

and t h a t

i s n o t a n a l g e b r a w i t h r e s p e c t t o pointwise o p e r a t i o n s .

In

o r d e r t o c o n s t r u c t a n a l g e b r a , we d e f i n e a n equivalence r e l a t i o n on

%

by saying

e q u i v a l e n t modulo the s e t o f a l l

z

n

V.

of

X

U

in

%.

r e l a t i o n on

f :U

+

written

X,

E U n V

W e denote by

f

P r o p o s i t i o n 1.

fx + Ex =

#(X)

U n V. Proof :

=

g(z)

X)

%,

are

t o mean t h a t

i s a neighborhood

f E

the quotient set of

ux,

we denote by

+

fx

. 5= Trivial.

the

X

is

i s a commutative a l g e b r a w i t h 1, i f

(fg)X,

+ g(z),

where

(f+g)(z) = f ( z )

where

(fg)(z) = f ( z ) g ( z ) ,

N

h f X = (XfIX,

by

~(x).

(f+gIX, lu

%

%-

cv

@

z E

f(z)

#(X)

if

in

i t s equivalence c l a s s i n

N

(modulo

both i n

The germ o f a holomorphic f u n c t i o n on

Definition 2 .

n V.

C,

It i s e a s i l y checked t h a t t h i s i s a n equivalence

equivalence c l a s s o f

z E U

f-g

where

t h i s equivalence r e l a t i o n :

we d e f i n e

-

g :V

and

(c

where

(Xf)(z) = Xf(z),

if

if

z c U.

if

79

LOCAL PROPERTIES

Remark 1. Each germ i n every point of

a l l over

X.

#(X)

has a well-defined value a t

However, tw germs i n

#(X)

may w e l l coincide

X . a n d be d i f f e r e n t germs (unless we add c o n d i t i o n s

x).

on

Notation. s h a l l denote by

If

R[[zl,..

. ,zn]]

s e r i e s with c o e f f i c i e n t s i n

1' *

we

the r i n g of a l l formal power

R,

Thus an element of

Yzn.

* *

1 # 0,

i s a commutative r i n g with

R

i n the indeterminates ( o r v a r i a b l e s )

. ..

R[ [ z ~ , ,zn]]

has a unique

express i o n

ml c al...ahzl - . - z c 4 n = 2 c za, n

z

a.>O

a

.

i = l , . . ,n 1-

al,...,an

20

n

Ul"

3;"

en

converge i n a neighborhood of

c j20

2,

%

t h e algebra of a l l formal

the subalgebra of

zn,

convergent power s e r i e s , t h a t i s , of a l l elements of

=

*

i n d e t e r m i n a t e s , with c o e f f i c i e n t s i n

We s h a l l a l s o denote by

Remark 2 .

and

E R

c

are integers.

For s h o r t , we s h a l l denote by power s e r i e s i n

where

a

Each

where

a E 3" aj

t h e j - t h component of

0.

We agree t h a t '3

of a l l

a;"

which

= Co = C .

can be w r i t t e n uniquely as

i s a j-homogeneous polynomial, c a l l e d a.

C.

LOCAL PROPERTIES

80

vj Proposition 2 .

r

If

=

aj;

( a ) because

every proper i d e a l i s contained i n some maximal i d e a l ( K r u l l ' s I n our c a s e , every proper i d e a l i s contained i n t h e

Theorem).

unique maximal i d e a l . ideal.

3.

I

i s equal t o the unique maximal

The r e s t of t h e statement has a l r e a d y been proved.

Definition 3 . unit

Hence

1 # 0, Then

I

A l o c a l r i n q i s a commutative r i n g

R

with

s a t i s f y i n g c o n d i t i o n ( a ) , ( b ) and ( c ) of Proposition

i s c a l l e d t h e u c a l of

Remark 3 .

I

For

R

and denoted by

R(R).

t o be an i d e a l , i t i s obviously necessary

and s u f f i c i e n t t h a t the sum of two n o n - i n v e r t i b l e elements be n o n - i n v e r t i b l e ; o r , e q u i v a l e n t l y , t h a t t h e sum of an i n v e r t i b l e element and a n o n - i n v e r t i b l e element be i n v e r t i b l e . Remark 4 . radical

R(R)

In a ring of

maximal i d e a l s of

R R.

R

with u n i t

1 # 0,

t h e Jacobson

i s defined a s the i n t e r s e c t i o n of a l l Thus, i n a l o c a l r i n g

R,

the s e t I = R(R).

LOCAL PROPERTIES

82

Remark 5 .

The i n t r o d u c t i o n of l o c a l r i n g s i s motivated by

the l o c a l study of a l g e b r a i c ( o r a n a l y t i c , o r d i f f e r e n t i a b l e ) v a r i e t i e s and f u n c t i o n s . Proposition 4 .

@"

and

3;"

a r e l o c a l algebras ( t h a t i s , The r a d i c a l of

a l g e b r a s which a r e l o c a l r i n g s ) .

of C?

= the s e t of a l l formal power s e r i e s whose c o n s t a n t term

n

i s zero.

en

and

3

Proof:

a r e a l s o i n t e g r a l domains. n

Case of

4 :

n

Each element

term i s obviously n o n - i n v e r t i b l e .

without c o n s t a n t

3

Conversely, each

c o n s t a n t term d i f f e r e n t from zero i s i n v e r t i b l e .

a

may assume t h a t the c o n s t a n t term of

a

by an a p p r o p r i a t e c o n s t a n t ) and w r i t e with c o n s t a n t term y = ~ + p +.

(p

3" = r a d i c a l

+

2 p

0.

=

with

I n f a c t , we

is

1 (by multiplying

= 1

-

p,

where

p E

zn

Consider then the formal power s e r i e s

..+++...

+ ...)

a

in

1. Thus,

n

men

3 .

a y = ( l + p + p

a i s i n v e r t i b l e and

-1 a

= y.

2

+...> We

have thus shown t h a t the s e t of a l l n o n - i n v e r t i b l e elements of

n

5

i s equal t o the set of a l l

Since t h e sum of two elements i n has zero c o n s t a n t term,

zn

a E

3;"

zn

with zero c o n s t a n t term. with zero c o n s t a n t term

i s a l o c a l a l g e b r a , with r a d i c a l

as i n d i c a t e d i n the s t a t e m e n t . Now,

zn

i s an i n t e g r a l domain:

Assume

a =

c a'

and

j 20 @ =

c J,'>O

4

where

aj

and

2

a r e j-homogeneous polynomials.

LOCAL PROPERTIES

a@ = 0 ,

Let

ah # 0 . ah Bo

+

h 1

a p

=

a

Thus,

," +

=

( ah @ 1+ah+l @o )

+ a.h+l po

- 0,

Let h+l a

+ ... etc.

,he1 + $+'Po

Next,

@l

a # 0.

and

+

= 0.

Now,

= 0

h

83

be the l e a s t i n t e g e r such t h a t

. ..

ah po = 0 ,

Hence

$po

. . ) ( p 0+p 1+. . .)

(,"+$+I+.

and

$8'

implies t h a t

B0

implies t h a t

0

=

=

0.

= 0.

Hence

and so on (using the f a c t t h a t polynomials form an

0,

i n t e g r a l domain). Case of

Cn:

We s h a l l use the isomorphism between

~ ( 0 ) (by Proposition 2 ) .

Then, one e a s i l y shows t h a t

non-invertible,

i s analytic a t

f(0) = 0.

where

f

conclude t h a t

0

Cn

i n the statement.

Next,

C"

0,

0 we

i s an i n t e g r a l domain, e i t h e r by

en

~ ( 0 ) o~r because

5n .

i s a subring,

(Actually, the case

n 3 , without using

could be handled a s the case of

Q.E.P.

Remark 6 .

en

is

i s a l o c a l algebra with r a d i c a l a s i n d i c a t e d

containing 1, of the i n t e g r a l domain

~(0)).

and

i f and only i f ,

i s holomorphic and vanishes a t

reasoning a s above using

Cn

. L e t t i n g

,

pj

. ,cnn

...,c n n > 4 .

SO

that

where

w1 -

= aj ( c L n y

aP ( c l n , .

in

... --

Since

. . ,cnn)

c

w i l l be d i f f e r e n t from z e r o . I n ' " ' "nn Next, we determine t h e c f o r k < n , so t h a t t h e matrix hk (chk) w i l l be i n v e r t i b l e . I f we do s o , we g e t an i n v e r t i b l e l i n e a r change of v a r i a b l e s , f o r which a series i n

wn

having the c o e f f i c i e n t

B(0,. Of

. . ,O,wn) Wg

w i l l be

d i f f e r e n t from

# 0,

LOCAL PROPERTIES

0

(4):

C

> 0,

X > 0,

and

such t h a t sup 1c;l K'

5CX~$,

f o r every

( 4 ) w i l l be proved by induction on we s a t i s f y ( 4 ) by r e q u i r i n g

C

1c:I.

j

j

,

and

0.

q.

For

j = q = 0,

Recall t h a t , f o r

j

L

1,

cq by using ( 2 ' )4j ; from which we g e t , by using j ( 4 ) inductively, f o r (zl,...,zn-l) E K' and X > 1 and p > 1:

we determined

91

LOCAL PROPERTIES

This w i l l g i v e us ( 4 ) i n d u c t i v e l y , i f we have

I n o r d e r t o guarantee t h i s , we s h a l l f i r s t assume t h a t

M < C. 0

l a r g e enough so t h a t

is

C

Next we s h a l l assume t h a t

lapl

[I,

-

i s l a r g e enough so t h a t

< C.

0

F i n a l l y , we

lapl assume t h a t Once

i s l a r g e enough s o t h a t

h

(5)

holds.

C , h , ~1 a r e s u b j e c t t o a l l i n d i c a t e d c o n d i t i o n s ,

Let u s use ( 4 ) t o

( 4 ) i s proved i n d u c t i v e l y a s shown above. prove t h a t

?.

y E

Consider t h e s e r i e s

c c q ( z l , . . . , z ~ )Zj - n. ~ J j ,420

By ( 4 ) , we see t h a t t h i s s e r i e s converges normally f o r lZll

xm E

j, and so Q.E.D.

S S

jo

j

j

. ..

c

i s an R-submodule of

and so i s generated by a f i n i t e number of elements

J

M

i s c l e a r l y non-empty,

h

If

h.

+ Rx;

be an R-submodule

N

R-submodules of

maximal i n S = S

Let

.

= S

".

My 9

xm

E

s.

S c S ; that is, j0 f o r a l l j 2 jo. Hence

Hence

jo

xl,

be

LOCAL PROPERTIES

96

M f

Proposition 2 .

Let

onto the R-module

N.

-1

(0)

and

M

Let

It i s obvious t h a t every

be noetherian.

i s noetherian.

R-module i s n o e t h e r i a n .

Moreover,

N

Hence

i s noetherian.

mapping a s s o c i a t i n g t o every subset of f

R-module

i s noetherian i f and only i f

M

Then,

R-submodule of a noetherian f-l(O)

be an R-homomorphism of the

are noetherian.

N

Proof:

f

I n f a c t , the

i t s inverse image under

N

induces an inclusion-preserving, one-to-one correspondence

between

R-submodules of

N

and

M

containing

be noetherian.

W e use the

R-submodules of

f-l(O). Conversely, l e t

f-'(O)

and

following easily-checked remark:

M

and i f

Let then

nf

P

So

n

n

(0) = Q

... c Sj

So

sequences

-1

f

-1

c

(0)

... be ... c Sj

and get

Let

f ( s j ) = f(S. ) S

j

= S

jo

a r e R-submodules of

P cQ

and

f ( P ) = f(Q),

R-submodules of

n f-l(O) c

then

M.

P = Q.

Then both

c and

*..

S. r l f - l ( O ) = S n f-l(O) J j0 jo. By t h e above remark, we

j o be such t h a t

for a l l

j

2

'0

,

for a l l

Corollary 1.

M 1 x . . .xM,

If

... c f ( S J. )

f(S0) c are stationary.

f-'(O)

N

If

j

jo.

M1,

...,Mn

a r e R-submodules, then

M

=

i s noetherian i f and only i f each f a c t o r i s . noetherian.

Proof: induction f o r

For

m = 2,

m > 2.

use the p r o j e c t i o n

M1xM2

--M2.

Use

LOCAL PROPERTIES

Corollary 2 .

Let

M

noetherian R-submodules,

97

be an R-module, and such t h a t

M1

+

M1

... + Mn

,...,Mn = M.

C

be

M

Then

M

i s noetherian. Proof:

Use the mapping

(xl

, .. . ,xn)

E M1x.. .xM n

- x1

-I-

. ..+ xn

E M

and the preceding Corollary and Proposition. Definition 2 .

A commutative r i n g

R

i s noetherian i f

R

i s a noetherian R-module. Remark 1. The above d e f i n i t i o n means of course t h a t

R

satisfies

the following equivalent c o n d i t i o n s : i).

Maximal Condition:

every non-empty s e t of i d e a l s of

R

has an element which i s maximal i n t h a t s e t . ii). ideals i n iii).

Ascending Chain Condition: R

every ascending sequence of

i s Sationary.

F i n i t e Generation Condition:

every i d e a l i n

R

is

f i n i t e l y generated. Comment : More g e n e r a l l y , we could consider a r b i t r a r y r i n g s with u n i t (not n e c e s s a r i l y commutative) and d e f i n e noetherian i f Example 2 .

R

R

t o be l e f t

i s a n o e t h e r i a n left-module. Every commutative r i n g i n which a l l i d e a l s a r e

p r i n c i p a l i s noetherian.

I n p a r t i c u l a r , the r i n g

i n t e g e r s , a commutative f i e l d i n one indeterminate

x

K,

and t h e , r i n g

with c o e f f i c i e n t s i n

K

Z

K[x]

of a l l

of a l l polynomials

a r e a l l noetherian

by t h e above. Remark 2 .

It f o l b w s from Proposition 2 t h a t i f

R

is a

LOCAL PROPERTIES

98

noetherian r i n g and

i s an i d e a l i n

I

then

R,

is a

R/I

Notice, however, t h a t i t i s n o t t r u e t h a t a

noetherian r i n g .

subring containing the u n i t of a noetherian r i n g i s n e c e s s a r i l y noe t h e r i a n . Proposition 3 . a noetherian r i n g Proof.

Let

Every f i n i t e l y generated R-module R

i s a noetherian module.

...xm

xl,

Rm

onto

M.

Since

neotherian module, so t h a t Proposition 2 .

Q .E .D

Proposition 4 .

be generators f o r

EM

R

f R

M

Proof:

R

(Hilbert).

w i l l be noetherian i f R

The r i n g

R[xl,..

x l , . . . ,xn

.,xn]

of a l l

with c o e f f i c i e n t s

R[x]


t h i s end, l e t

I

be an i d e a l i n


R[x]

.

... .

i s an i d e a l i n

i s noetherian. For an i n t e g e r

R

i n a polynomial i n

I(k)

i s noetherian.

We s t a r t with the case

R

c o n s i s t of a l l elements i n

xk

. ,xn]

R.

of a l l polynomials i n one

x

c o e f f i c i e n t s of

R[x l , . . . , ~ n ] onto

R[xl,..

indeterminate

I(k) c

is a

i s noetherian.

R

be n o e t h e r i a n .

n = 1 and prove t h a t the r i n g

I(k)

Rm

M

The mapping a s s o c i a t i n g t o each polynomial i t s

Conversely, l e t

let

+ hmXm E

.

constant term i s a r i n g homomorphism of Hence

+ ...

The

i s a noetherian module, by

i s noetherian i f and only i f

R

1x 1

M.

i s a noetherian r i n g ,

polynomials i n the indeterminates in

-X

m

(xl,.. . , l a )

R-module homomorphism maps

over

M

k

2

0,

appearing a s t h e I

R

To

of degree and t h a t

5 k. I(0) c

...

99

LOCAL PROPERTIES

W e claim t h a t i f

I

J

f o r all

k

20

I(k) = J(k),

a r e two i d e a l s i n

I = J.

then

be of degree

5

k.

Then t h e r e i s some

such t h a t

-

fk

i s of degree

g

-

fk E J.

5k

-

1, such t h a t

see t h a t degree

g

Continuing, we g e t

g

hence

f o -t-

=

fk,...,fo

.. . + f k

E I.

Let u s now prove t h a t

...

I. c J Ij(k), with

be i d e a l s i n

i n R, or

j

k

k.

Since

R

j

~ . ( k )= I. (ko), J Jo k 2 ko, we have

-

fk

for j

k,

j

L joy

k

'Y

ko.

-

g

5

-

k

- ... -

fk

i s noetherian.

Let

2.

fo = 0 ;

I.

c

... c

Consider t h e corresponding i d e a l s Notice t h a t

I.(k) J

increases

i s n o e t h e r i a n , t h e r e are

Since

Of

i s of degree

k

L

ko.

Ij ( k ) .

j o y ko

In p a r t i c u l a r ,

I n f a c t , if j

I j ( k ) = Ij (ko) = I. ( k ) .

2

we

Hence,

? j o y

I j ( k ) = Ij (k),

JO

R

0

i s n o e t h e r i a n , f o r every f i x e d

the sequence Io(k) c

(1)

i s stationary. j

fk-l

1. k ,

I = J.

Therefore

0

for

I c J,

E 1 such t h a t

0.

2 joy

g E J

of degree

Since

fk-l

i s maximal among a l l

I. (k ) Jo O

such t h a t

-

R[x]

R[x].

,

for

1.

Then t h e r e i s some g

and

In f i c t , l e t

fk E I ,

-

2k

R[x]

2

. .. c I j ( k )

c

.. .

In f a c t , we j u s t saw t h a t i t i s s t a t i o n a r y f o r

j o y provided

k

ko.

t h e f i n i t e l y many cases

0

5k

.

cc.6 (k = 0 , . j=1J ~ Y J

.. ,p-1).

Then we g e t , from (l),

-

c, -h

This shows t h a t

S

c ~ c=j p j y~ where c

I

= y

-

S

c

c

j=1 j

i s f i n i t e l y generated, namely by

F i n a l l y , l e t us take c a r e of the case i n which s a t i s f y any condition

(Cp).

Since

a # 0,

l i n e a r change of v a r i a b l e s , the image of (C 1 P i s the image of

condition

(015, Proposition 2 ) . I

(resp., ?).

B~,...,$,. does n o t

by an i n v e r t i b l e

w i l l s a t i s f y some

Therefore the id.eal which

under t h i s change of v a r i a b l e s w i l l b e

f i n i t e l y generated. generated.

a

zn

y. E J

I t follows t h a t

I

itself is finitely

Q .E .D.

Remark 3 .

If

U c Cn

i s not a noetherian r i n g .

u ( ~= ) @",f o r

77 E

any

i s open and non-void, then

w(U)

This i s i n c o n t r a s t with the f a c t t h a t

i s a noetherian r i n g .

L e t us show t h i s

LOCAL PROPERTIES

102

i n case

U

We use the f a c t (due t o Weierstrass) t h a t ,

= (c.

TI^,...,^,...

given a sequence numbers tending to

,

OD

o f pairwise d i s t i n c t complex

there i s

#(a)

f E

vanishing e x a c t l y

a t these p o i n t s . formed by a l l

Consider the i d e a l I in #(C)

#(a)

f E

vanishing a t a l l b u t a f i n i t e number of these p o i n t s . f i n i t e number of elements such t h a t a l l

f

fly...,fs

( j = l,...,~)

a t a l l points of t h e sequence

There i s some

I.

.

vanish a t

j Weierstrass Product Theorem, t h e r e i s ql,..

Fix any

nm

0

By the

0

#(a) vanishing e x a c t l y

f E

. ,%, . . .

except a t

. . . gSfS) %O

Then

f E I,

with a l l

but

f

cannot be w r i t t e n

g . E #((I), for f ( x ) J 0

a s an i d e a l .

Hence

Remark 4 . and only i f

zn

z

m

{O')

, @n)

(resp.

be an i d e a l .

(m = 0,l ,...).

am # 0 .

invertible since

The i d e a l

In f a c t , l e t I

{O)

m

I

m

i s principal.

Let

i s generated by some

be t h e l e a s t p o s i t i v e a = a zm m

c o n t a i n s some

+ am+lzm+l+. . .

i s contained i n the p r i n c i p a l

I

z ,

Moreover,

a # 0; m

hence

p r i n c i p a l i d e a l generated by I

I

i s a principal i d e a l ring i f

We claim that

Clearly,

i d e a l generated by

that

generates

n = 1.

i n t e g e r such t h a t with

I

= 0

i s not a noetherian r i n g .

~((c)

In f a c t , l e t n = 1.

I

# 0 and a l l f j ( % o )

Thus, no f i n i t e subset of

( j = l,...,~).

+

f = glfl

z

m

B

=

am + amflz

zm = up-'

E I.

z

m

is

Hence, t h e

i s contained i n

i s the p r i n c i p a l i d e a l generated by

+ .. .

.

I.

It follows

LOCAL PROPERTIES

Let now

n

2 2.

10 3

generated by a l l s e r i e s without constant terms. principal. Inded,let any

p

=

po

+ p1 +

a E I.

.. . i n

zn(resp.,

ap

I

a = al 4- a2 4-

Write

2'1,

higher order homogeneous polynomials. s c a l a r m u l t i p l e of

I,

Consider the unique maximal i d e a l

we have

Since

peal

i s not

.. . . aB

by a s i n g l e element).

22

- $,a1

is a

al, we cannot g e t a l l elements o f

(since a v e c t o r space o f dimension

For

I

as

cannot be spanned

+

LOCAL PROPERTIES

104

$17. Unique F a c t o r i z a t i o n P r o p e r t i e s We s h a l l consider an i n t e g r a l domain

1 # 0,

commutative r i n g with cancellation l a w holds.

if x

x, y E

* R

divides

y

of

y.

R

x-y E

i f there i s

t E

xly

be t h e m u l t i p l i c a t i v e s e m i -

d i f f e r e n t from

* R ).

then

we then w r i t e

i n which t h e m u l t i p l i c a t i v e

* R

Let

group of a l l elements i n

that i s , a

R;

If

* R

and say t h a t

x

x, y E

0

* R ,

(notice t h a t we say t h a t

(unique) such that

y = tx;

i s a factor o r divisor

This r e l a t i o n has t h e p r o p e r t i e s : (i)

x Ix

(ii)

if

(reflexivity) x ( y and xIz

(iii)

xly

ylz,

then

(transitivity).

iff

x.zIy.z,

for all

(Compatibility

z.

with m u l t i p l i c a t i o n ) . Clearly,

x

invertible.

i s a f a c t o r of every element i f and only i f x

and

y

a r e associated i f

i s the c a s e i f and only i f write x-y

x-y iff

y = tx

with

xly t

and

ylx.

invertible.

is

x

This We

and thus g e t an equivalence r e l a t i o n such t h a t x-z-y.z

for a l l

z

( c o m p a t i b i l i t y with m u l t i p l i c a t i o n ) .

The i n v e r t i b l e elements a r e those a s s o c i a t e d t o 1. A d i v i s o r

of

y

i s proper i f

to

y.

Otherwise,

x x

i s n e i t h e r i n v e r t i b l e nor a s s o c i a t e d i s a t r i v i a l d i v i s o r of

i r r e d u c i b l e o r extremal i f proper d i v i s o r . R

generated by

x

y.

x

is

i s n o n - i n v e r t i b l e and has no

This means t h a t t h e p r i n c i p a l i d e a l x

x

Rx

in

i s maximal among a l l proper p r i n c i p a l i d e a l s

LOCAL PROPERTIES

of

i t does n o t mean t h a t

R;

Notice t h a t i f

x E

* R

i s a maximal i d e a l .

Rx

i s written

product of i r r e d u c i b l e f a c t o r s , then another expression of where

u

x

x = t 1 * - .nt as a

x = t a ( ~ ) * . ' ~ u ( ni > s

as a product o f i r r e d u c i b l e f a c t o r s ,

{l, ...,n ] .

i s a permutation of a r e i n v e r t i b l e and

u l ' " ' yun

105

Moreover, i f

u l - . - u n = 1,

i s another expression of

(ul*tl)---(un-t ) n

then

x =

as a product

x

of i r r e d u c i b l e f a c t o r s . A unique f a c t o r i z a t i o n domain i s an i n t e g r a l domain having the following two p r o p e r t i e s :

D1) : x

Every n o n - i n v e r t i b l e

*

x E R

R

may be w r i t t e n

t l * * * t as a product of a f i n i t e number of i r r e d u c i b l e f a c t o r s .

n

D2) :

Every n o n - i n v e r t i b l e

x E R

*

has an e s s e n t i a l l y unique

expression as a product of a f i n i t e number of i r r e d u c i b l e f a c t o r s , i n the sense t h a t , i f expressions,

El,.

. . ,n7 x E

whenever

n = r

then

such t h a t

* R

x = t

1

- - tn '

- - - sr

a r e two such

and t h e r e i s a permutation

a

of

si& t o ( i ) .

i s s a i d t o be prime i f

x Iyz,

s1

=

then n e c e s s a r i l y

t h a t the principal i d e a l

Rx

in

x x Iy

i s n o n - i n v e r t i b l e and i f or

xIz

.

generated by

R

This means

x

i s a prime

idea 1. Proposition 1. Every prime element i s i r r e d u c i b l e . Proof: Since

Let

x](y.z),

*

x E R

be prime, and assume t h a t

then e i t h e r

xIy

or

xlz.

x = y-z.

Assume that

xly.

LOCAL PROPERTIES

106

Then z

that i s ,

y = tx;

i s invertible.

Hence

x = txz

Similarly, i f

i s irreducible.

x

and so xlz,

tz

1.

=

then

Therefore

i s invertible.

y

Q.E.D.

We introduce now t h e c o n d i t i o n : Every i r r e d u c i b l e element i s prime.

D2’):

Proposition 2 . conditions

D2)

I n an i n t e g r a l domain s a t i s f y i n g

and

D2‘)

a r e equivalent.

L e t u s show t h a t

Proof:

D2)->

i r r e d u c i b l e and n o n - i n v e r t i b l e . y-z = t.x, hence

t.

f o r some

xlz.

If

Similarly, i f

then t h a t n e i t h e r

y

nor

i n v e r t i b l e ; f o r otherwise

Dl),

D2’).

Consider

XI

Assume t h a t

i s i n v e r t i b l e , then

z

i s invertible. -1

z

x = (t

y)z

and

z = (y -It )x ;

xly.

Then

x

*,

that i s

(y-z);

i s i n v e r t i b l e , then

y

x E R

t

Assume

i s not

would be t h e

product of two f a c t o r s n e i t h e r of which i s i n v e r t i b l e , c o n t r a d i c t i n g the assumption t h a t

i s irreducible.

x

t = t l - - - t n ,y = yl.-.yp,

write

are irreducible. either

xzyi

either

xly

Hence , y1

f o r some or

x(z.

Now we show t h a t

i,

or

Hence, D2’) =>

on the number of f a c t o r s .

- - - z q --

xdz x

Call

Dl),

j

we can

where a l l f a c t o r s

= zl”’zq

yp z1

* * *

By

t

...~tnX .

f o r some

j;

i s prime, and so D2). D2),

By

D2),

that is, D2‘)

holds.

We proceed by induction the f a c t t h a t

D2)

is

t r u e f o r every n o n - i n v e r t i b l e element which may be w r i t t e n as a product of

n

irreducible factors.

LOCAL PROPERTIES

D2)1

i s true.

In fact, let

x

107

be i r r e d u c i b l e , and assume

.

x = xl-*-x with xi i r r e d u c i b l e ( i = 1 , . . , p ) . By P' D2'), x i s prime. Since x l x l . . . x i t follows t h a t x P' d i v i d e s some x say x l ; hence xuxl ( s i n c e both a r e i' that

That i s

irreducible). Then

uxl

then

x2,

x = ux

1 7

where

u

i s invertible.

implies t h a t If P 2 2 , = x2"'xP' P would be i n v e r t i b l e , b u t t h i s i s n o t p o s s i b l e

= x l . . .x

'xP s i n c e they a r e i r r e d u c i b l e . * *

Now, assume Let

x

D2),

and x ' i j since x n+l

p

=

1 and

and l e t us show t h a t

D2)1

D2)n+l

holds. holds.

= x i . . .XI where a l l n+l P' a r e i r r e d u c i b l e . Since xn+l I ( x i . ex' ) and P i s prime (by D 2 ' ) ) , we have t h a t xnfl d i v i d e s

be n o n - i n v e r t i b l e ,

x

Hence

x

=

xl..

'X

and x ' a r e a s s o c i a t e d Therefore x x' say X I . j' P n+l P where u s i n c e both a r e i r r e d u c i b l e . That i s , x ' = ux n+l ' P i s i n v e r t i b l e . W e then g e t x l - - ~ ~ - xx i .~* . ~ +k~ - ~ ( u x ~ +o~r ) , some

. . .xn

x1

= (uxi).

.

.XI

p-1'

We then have t h a t a l l f a c t o r s i n the l a s t equation are irreducible.

By

D2)n,

t h e two above decompositions a r e

e s s e n t i a l l y t h e same, from which i t follows t h a t t h e two decompositions f o r holds.

x

are e s s e n t i a l l y t h e same.

Hence

D2)n+l

Q .E .D.

Let us now give another v e r s i o n of uniqueness of f a c t o r i z a t i o n by using g r e a t e s t common d i v i s o r s .

LOCAL PROPERTIES

108

*,

x,y,z E R

and

z,

written

and

z,

and every element d i v i d i n g both

divide

x

i s a g r e a t e s t common d i v i s o r of

If

x = gcd ( y , z ) = y

A

z,

y

if

x

y

and

z

must

y

and

z

must be

y

d i v i d e s both

x.

Any two g r e a t e s t common d i v i s o r s of

a s s o c i a t e d ; any element a s s o c i a t e d t o a g r e a t e s t common d i v i s o r of

y

and

i s a l s o a g r e a t e s t common d i v i s o r of

z

'hm elements a r e r e l a t i v e l y prime i f

y

and

z.

1 i s a greatest

common d i v i s o r of them. Let u s introduce the c o n d i t i o n D2"):

Any two elements i n Remark 1.

R

*

have a g r e a t e s t common d i v i s o r .

The concept of g r e a t e s t common d i v i s o r extends

i n an obvious way t o f i n i t e l y many elements, and then implies t h a t any f i n i t e number of elements i n

* R

D2")

has a g r e a t e s t

cormnon d i v i s o r . The d u a l concept of a l e a s t common m u l t i p l e may

Remark 2 . be defined.

I t s e x i s t e n c e amounts t o t h a t of a g r e a t e s t common

d i v i s o r , s i n c e the product of the l e a s t common m u l t i p l e and the g r e a t e s t common d i v i s o r of two elements i s equal t o t h e product of these two elements. Lemma 1.

Under

a)

x = x

b)

x

C)

X'(Y

d)

if

A

A

D2"),

we have

y

xly

iff

(y AZ)'(X A Z>

x

A

A

y)

(XY)

y = 1, x

A

z

(Xz)

A

A

z = 1,

then

x

A

(yz)

=

1.

LOCAL PROPERTIES

Proof:

a)

and

b)

To prove c ) , l e t

vly

and

vIz

w = txv. txvIxy, tvlv.

w

and

Conversely

wlxy,

wlxz.

-

b) Y

A

and so

d),

x

Hence

Then

xv(w.

Hence

By s u b s t i t u t i o n , we g e t

tll;

tvly

and

hence

y = 1 implies

A

(zx)

(zy)] = 1. x

A

A

(zy) = z .

Thus

zy = 1,

conditions Proof:

(xz).

A

tvlz.

Hence

i s invertible

t

xv.

[x

since

z(x

Hence

A

(zx)]

x

A

y) * z ,

A

x

which

z = 1

A

implies

( z y ) = 1, by p a r t

A

(zx) = x .

In an i n t e g r a l domain s a t i s f y i n g c o n d i t i o n

Proposition 3 . Dl),

xvlxz.

and by c a n c e l l a t i o n

by p a r t c ) implies [(zx)

w = (xy)

z,

A

xvlxy

To prove

A

v = y

By c a n c e l l a t i o n a g a i n ,

and so

x

a r e immediate.

imply

txvlxz,

109

D2),

D2')

D 2 ) => D2").

f a c t o r i z a t i o n domain.

If

and

a r e a l l equivalent.

D2")

In f a c t ,

R

i s then a unique

i s a n o n - i n v e r t i b l e element i n

x

R

Jr

,

by w r i t i n g i t as a product of i r r e d u c i b l e f a c t o r s and c o l l e c t i n g together a s s o c i a t e d f a c t o r s , we g e t x

i s i n v e r t i b l e , pi

utlP1.

- . tn Pn ,

a r e p o s i t i v e i n t e g e r s , and

non-associated i r r e d u c i b l e elements. write

=

If

t

t x = u t P 1 - - -pn y = v t q l - - - qn, t 1 1 n n y

y E R

where

u

invertible ,

*,

and

and q non-negative i n t e g e r s , and t 'j j j mutually non-associated i r r e d u c i b l e elements. Then i t i s immediate t h a t

xly

iff

pj

5 qj

( j = 1,.

. ., n ) .

x

and

y,

z = t r l * - * tn ,

1

n

we may v are

It follows

r

t h a t , f o r any such

u

a r e mutually

j

and

x

where

where

are

.LOCAL PROPERTIES

110

r

= inf

j

Hence,

(p q ), i s a g r e a t e s t common d i v i s o r o f j yj D2") holds.

Now, xlyz. (x

A

By

=>

D2") D2"),

y)lx,

and

D2').

x x

A

y

In fact, l e t

x

and

A

z

x

x

then

and

y.

be i r r e d u c i b l e and

Now s i n c e

exist.

i s irreducible,

x

A

is either

y

i s either

i n v e r t i b l e o r i s associated to

x.

Similarly,

i n v e r t i b l e o r i s associated t o

x.

We c l a i m t h a t i t i s n o t

possible t o have both

is x

x A

y = 1 and

A

(yz) = 1.

But

x

A

y

x

A z =

x

A

and

Hence

x

x

A

y = x

or

i s prime and

D2') =>

D2)

x

(yz) = x ,

A

A

z

A

invertible;

z

that

1. For then, by Lemma l ( d ) ,

x r 1, ~ which i s impossible s i n c e either

x

x

and then we conclude t h a t x

i s irreducible. that i s ,

z = x;

xly

or

Hence, xlz.

D2') holds.

w a s proved i n Proposition 2 .

Q.E.D.

Let u s introduce n e x t t h e following condition :

a;): t h a t each

f a i l s t o c o n t a i n a sequence

R*

xj+l

Remark 3 . that

D1')

i s a proper f a c t o r of

x

j

xl,..

.

,Xj

,.. .

such

.

I n the language o f i d e a l theory, i t i s immediate

means t h a t every ascending sequence of p r i n c i p a l i d e a l s

i s s t a t i o n a r y from a c e r t a i n index on, which i s t h e ascending chain condition f o r p r i n c i p a l i d e a l s .

This i s e q u i v a l e n t t o t h e f a c t

t h a t every non-empty s e t of p r i n c i p a l i d e a l s of

R

has an

element which i s maximal i n t h a t s e t , which i s t h e maximal condition f o r p r i n c i p a l i d e a l s .

LOCAL PROPERTIES

Proposition 4 .

and D1')

or

D2')

s a t i s f y i n g any one

R

D2"),

the conditions

D1)

Assume

D2), D2')

and any one of

x E R

because each n o n - i n v e r t i b l e

*

D2").

or

D1') i s c l e a r ,

i s a unique f a c t o r i z a t i o n domain and

R

D1)

a r e equivalent.

Proof: Then

I n an i n t e g r a l domain

D2),

of t h e c o n d i t i o n s

111

has only f i n i t e l y many non-

mutually a s s o c i a t e d d i v i s o r s .

Dl')

Conversely, we s h a l l prove t h a t (without using c o n d i t i o n s be n o n - i n v e r t i b l e . If

D2), D2') o r

We claim t h a t

D2")).

i t has a proper f a c t o r through. and so on.

Dl'),

By

many s t e p s .

i s i r r e d u c i b l e , we a r e

x

be n o n - i n v e r t i b l e .

Write

If

x. Assume, then, t h a t x

= t y

1 1'

where

tl

x,

If

y1

so assume t h a t

where

i s i r r e d u c i b l e , then yl

i s not i r r e d u c i b l e .

t2 i s i r r e d u c i b l e , and s o on.

i s not

x

i s irreducible Then

D1)

y1

verified for

x.

Dl').

i s non-

i s true f o r

Write

Y 1 = t2Y2,

This process m u s t s t o p

a f t e r f i n i t e l y many s t e p s , f o r otherwise t h e sequence would c o n t r a d i c t

x.

i s irreducible,

x

(which can be done by the above paragraph). invertible.

x2

t h i s process has t o s t o p a f t e r f i n i t e l y

D1) holds f o r

irreducible.

x1

Then we w i l l s t o p a t some i r r e d u c i b l e f a c t o r of

Now, l e t then

If

*

i s not i r r e d u c i b l e ,

x

If

i s not i r r e d u c i b l e , i t has proper f a c t o r

x1

If

xl.

x E R

Let

has an i r r e d u c i b l e d i v i s o r .

x

i s i r r e d u c i b l e , we a r e through.

x

Dl),

implies

When we a r e forced t o s t o p ,

yl, . . . , y . , . .

J

D1) i s

LOCAL PROPERTIES

112

Proposition 5 .

Every p r i n c i p a l i d e a l domain i s a unique

f a c t o r i z a t i o n domain. Proof:

Let us prove

(and hence

Dl')

Dl))

i n the form of

the ascending chain c o n d i t i o n f o r ( p r i n c i p a l ) i d e a l s . if

...

In f a c t ,

...

c a r e ( p r i n c i p a l ) i d e a l s , then I = U I j j j i s an i d e a l , hence the p r i n c i p a l i d e a l generated by some

I

1

x E I. I

j

c

Let

= I

jo

c I

x F I , j0

for

j 2j

0

f o r some

0

.

Dl')

So

exists

z E R

we g e t

z Ix.

x

and

some

y

y.

Rx

+ Ry

Rx

Similarly

a,b E R.

Then i t i s c l e a r t h a t

holds.

is satisfied.

generated by

such t h a t

Moreover,

.

D2")

L e t us now v e r i f y t h a t Take t h e i d e a l

j

z ly.

+ Ry

= Rz.

Hence

z E Rz = Rx

x

z

+ Ry;

z.

Hence

Example 1. The r i n g

Since

2

x,y E R

Now, t h e r e

x 6 Rx c Rz,

hence

z = ax

and so

D2")

+ by x

x

for and

holds.

of a l l i n t e g e r s and t h e r i n g

K[x]


a r e known t o be p r i n c i p a l i d e a l domains.

proved using the Euclidean d i v i s i o n a l g o r i t h m ) . a r e unique f a c t o r i z a t i o n domains.

*.

i s a common d i v i s o r of

z = x A y,

of a l l polynomials i n one indeterminate K

y.

It follows t h a t every common d i v i s o r of

i s a d i v i s o r of

in a field

and

Let

(This i s

Hence, they

More g e n e r a l l y , t h e r e i s

t h e n o t i o n of a Euclidean domain, which i s always a p r i n c i p a l i d e a l domain and hence a unique f a c t o r i z a t i o n domain.

LOCAL PROPERTIES

Proposition 6 .

1 # 0.

(Gauss).

Then the r i n g

inde t e rmina t e s

,xn

be a commutative r i n g with

R

.. . ,xn]

R[xl,

-

XI'.

Let

113

of a l l polynomials i n t h e


f a c t o r i z a t i o n domain i f and only i f

R

i s a unique

i s a unique f a c t o r i z a t i o n

R

domain. Proposition 6 , we need the following

Before we prove preliminaries: p E R[x]

,

p

Assume

# 0.

i s a unique f a c t o r i z a t i o n domain.

R

W e say t h a t

p

i s primitive i f

common d i v i s o r of a l l i t s non-zero c o e f f i c i e n t s .

Let

1 i s a greatest A t any r a t e , i f

r = the g r e a t e s t common d i v i s o r of i t s non-zero c o e f f i c i e n t s , then we may w r i t e

where

p = r p ' E R[x]

i f we have p'

p = rp',

,

r

i s p r i m i t i v e , then

pl

i s primitive.

E R[x]

r E

where

* R ,

pl

Conversely,

E R[x] , where

i s a g r e a t e s t common d i v i s o r of a l l

non-zero c o e f f i c i e n t s of

Since t h e g r e a t e s t common d i v i s o r

p.

i s unique up t o a s s o c i a t i o n , i t follows t h a t t h e decomposition p = rp'

,

with

*,

r E R

p ' E R[x]

,

primitive,

t o a s s o c i a t i o n by an i n v e r t i b l e element i n A content of

p E R[x]

,

p

i s unique up

R.

# 0, i s a g r e a t e s t common

d i v i s o r of i t s non-zero c o e f f i c i e n t s .

A content i s umue up t o

association. Lemma 2 (Gauss).

Let

R

be a unique f a c t o r i z a t i o n domain.

The product of two p r i m i t i v e polynomials i s p r i m i t i v e .

The

product of the c o n t e n t s of two non-zero polynomials i s the content of t h e product of these polynomials.

LOCAL PROPERTIES

114

Proof:

Let

p , q E R[x]

i s not primitive.

0

Since

t E R*

d i v i d i n g a l l non-zero c o e f f i c i e n t s of

i s primitive,

p

u

Call

p.

t h e l e a s t i n t e g e r such t h a t

a

i s n o t d i v i s i b l e by t . S i m i l a r l y , l e t v c1 i n t e g e r such t h a t b v i s n o t d i v i s i b l e by t . x tr+v

in

pq

pq.

cannot be a common d i v i s o r of all

t

non-zero c o e f f i c i e n t s of

looks l i k e

c

= lL+V

terms preceding o r following

a b

and

i s n o t p r i m i t i v e , t h e r e i s an i r r e d u c i b l e ,

pq

hence prime,

of

pq

+ . . . + bnxn.

q ( x ) = bo

Since

+ . . . + amxm

p(x) = a

Write

Assume t h a t

be p r i m i t i v e .

i s not d i v i s i b l e by

a b

/Jv

be the l e a s t The c o e f f i c i e n t

... + a Pb V + . . . .

a r e d i v i s i b l e by

t,

All but

But t h i s c o n t r a d i c t s the assumption

t.

P V

that

i s d i v i s i b l e by

c u+v

p = rp',

Next, i f we w r i t e p',q'

E R[x],

Let Then

K

primitive,

rs

and hence

i s a content f o r

contains

Lemma 3 .

R

If

is irreducible i n Proof:

of

p(x)

then

Hence

pq

q = sq',

If

and

Q.E.D.

pq.

invertible i n

contains

i t s quotient f i e l d . R[x].

p(x)

p'(x) R[x]

K

R[x], p(x)

,

and

p(x)

is irreducible i n

i s irreducible i n p(x) = r p ' ( x ) ,

R[x]

,

where

i s p r i m i t i v e , we have t h a t s i n c e degree

i s primitive

p'q'

i s of degree > 0

then

r , s E R,

where

p(x) E R[x]

For, writing and

K[x]

i s primitive.

pq = r s p ' q ' ,

be an i n t e g r a l domain and

R

primitive.

t.

p > 0

K[x].

then i t i s

i s a content

r p'

i s not

and hence degree p ' > 0 .

LOCAL PROPERTIES

r

Hence

is invertible i n

,

R[x]

115

and so i n

R.

Hence

p

i s primitive. Now, assume q(x) n(x),

where

have degrees > 0 . q ' ( x ) E R[x] written

,

i s reducible i n

p(x)

W

q(x)

i s primitive.

q'

where

p(x)

k

c

hq'(x),

where

Similarly,

K,

Write

h E K,

n(x)

n ' ( x ) E R[x]

hk q ' ( x ) n ' ( x ) .

Therefore

may be

and

hk =

n'

W'

wp(x) = v q ' ( x ) n ' ( x ) .

may be w r i t t e n a s a product of

R[x]

and a p r i m i t i v e polynomial i n an e s s e n t i a l l y

R

unique way, we see t h a t u =

=

=

Since a polynomial i n

hence

p(x>

We may w r i t e

Then

an element i n

so t h a t

and a r e n o t i n v e r t i b l e and

n(x) = kn'(x),

where v , w E R .

,

q ( x ) , n ( x ) E K[x]

and

i s primitive.

K[x]

v

w must be a s s o c i a t e d i n

and

i s an i n v e r t i b l e element i n

p(x) = [ u q ' ( x ) ]

R.

i s f a c t o r i z a t i o n of

n'(x)

R,

Therefore, p(x)

in

R[x]

i n t o n o n - i n v e r t i b l e f a c t o r s , c o n t r a d i c t i n g t h e assumption t h a t p(x)

i s irreducible in

R[x]

Proof of Gauss' Theorem: Assume Notice t h a t

R[xl,..

. ,xn]

.

Q.E.D.

Proposition 6 :


i s isomorphic t o t h e subring of

R

R[xl,..

. ,xn]

of a l l c o n s t a n t polynomials, and t h a t an i n v e r t i b l e polynomial must be c o n s t a n t .

Then i t follows t h a t

R

i s a unique

f a c t o r i z a t i o n domain. Conversely, l e t induction, i f

n

L

2,

R

be a unique f a c t o r i z a t i o n domain.

and s i n c e

R[xl,.

. . ,xn ]

= R[xl,.. .,

By x ~ - [x,] ~ ]

,

LOCAL PROPERTIES

116

i t i s s u f f i c i e n t to prove t h e proposition f o r i t i s enough t o show t h a t

n = 1.

That i s ,


R[x]

We s t a r t by showing t h a t t h e r e e x i s t s a f a c t o r i z a t i o n i n t o irreducible factors. where

h F R,

and

E R[x]

in

R[x]

,

p'(x)

qi

p' of A l s o , t h e degree,,,;

0.

Write If

i s not

p'

It follows t h a t

qi

i s not a

would be i n v e r t i b l e i n

R,

hence

qi

Moreover,

i s not irreducible i n

R[x]

q i > 0.

Thus, t h e degree of and

q;

a r e primitive

s i n c e they a r e f a c t o r s of a p r i m i t i v e polynomial. q i , q;

p(x) = h p ' ( x ) ,

where

i s primitive.

since

# 0.

p

q i ( x ) q;(x),

=

a r e not i n v e r t i b l e .

c o n s t a n t , f o r otherwise

,

i s primitive.

p ' ( x ) c R[x]

R[x] , w r i t e

irreducible i n

q i , q;

p(x) E R[x]

Let

,

I f one of

we can r e p e a t t h i s process.

This cannot be repeated more than a number of times equal t o the degree

p ' = degree p ,

degree > 0 .

because each t i m e each f a c t o r has

Therefore, we g e t a decomposition of

f i n i t e number of f a c t o r s i r r e d u c i b l e i n Write get

h = hl...hm, p = hl...h

R[x]

r ' .r' m 1" t

and assume

R[x]

ri.. . rt' '

Then we into

.

p(x) = h l . . . h

a r e two decompositions of

p

r'(x) m l

i s primitive.

Take

...r;(x)

p(x) E R[x]

= k, ...-;(

into irreducible factors.

have seen that every polynomial i n of degree > 0

p' =

h a r e irreducible i n R. j w i l l be a decomposition of p

L e t us now prove e s s e n t i a l uniqueness.

# 0,

say

where

irreducible f a c t o r s i n

p

,

into a

p'

R[x]

x)

,

...s i ( x >

We

which i s i r r e d u c i b l e and

(See the proof of Lemma 3 ) .

Also,

a polynomial may be w r i t t e n as the product of an element i n

R

LOCAL PROPERTIES

117

Hence

and a p r i m i t i v e polynomial i n a unique way.

and without loss of g e n e r a l i t y , we may assume

i s a unique f a c t o r i z a t i o n domain,

m = M

a s u i t a b l e rearrangement of t h e o r d e r .

. .rL(x)

ri(x).

= si(x).

..s$(x).

-

kl..

.%,

...%

hl...h

= k m 1 s i ( x ) . . . s $ ( x ) ) . Since

(by including an i n v e r t i b l e element with R

h l . . .h m

and

hiw ki

after

Then, by d i v i s i o n ,

Since

R[x] c K[x]

and

K[x]

i s a unique f a c t o r i z a t i o n domain and t h e f a c t o r s i r r e d u c i b l e i n R[x]

must be i r r e d u c i b l e i n

(by Lemma 3 ) ,

K[x]

and a f t e r rearrangement of t h e o r d e r ,

ri(x)

r

t = T

then

s;(x)

in

K[x]

.

V

That i s

s' = c.r'

i

vi, wi

where

iy

1

E R.

where

ci E K ,

wi s i'

Then

r'

a r e p r i m i t i v e , we g e t t h a t

in

R,

i

and s o

r'- s ' i

in

i

=

c

v r' ii'

v.r" w 1

R[x].

# 0.

i

Write

and s i n c e Thus

i'

ci

ci = s'

i

i w, i

and

i s invertible

Q.E.D.

We use the above r e s u l t s t o prove:

$'

Proposition 7.

(resp.,

2')


domain. Before we prove Proposition 7 , l e t us e s t a b l i s h t h e following remarks :

1) The product of a f i n i t e number of d i s t i n g u i s h e d polynomials in

zn

(resp.,

Cn)

i s a d i s t i n g u i s h e d polynomial.

(See $15,

Definition 2 ) .

2)

I f a d i s t i n g u i s h e d polynomial i n

f i n i t e product o f in

n-1

3;

(resp.

polynomials.

,

monic polynomials i n @n-1 ),

zn zn

(resp.,

cn)

is a


then these f a c t o r s a r e d i s t i n g u i s h e d

LOCAL PROPERTIES

118

To justify this, it is enough to consider the product of two monic polynomials.

So,

we have

.

(ao+alzn+. .+z~)(bOtblzn+..+zq) n

a distinguished polynomial. Consider the least i is invertible; of course, this

i exists and is

=

such ai

5 p (ap

=

1).

A l s o , consider the least j

of course j 5 q.

such that b is invertible; j: It then follows that the coefficient of

i* is invertible and the remaining terms are non-invertible. n Since the product polynomial is distinguished, we conclude that

Z

i+j=p+q, a

o

+ aIz n +

PO lynomiaIs

3)

.

and that i

. . . + zpn

=

and bo

p

and j

=

q.

Therefore

+ blzn + . . . + zqn

are distinguished

If f = p.u are Weierstrass decompositions of j ~j l,...,n (that is, decompositions in the sense of the

j= f jy Weierstrass Preparation Theorem, 015, Theorem l),

then

T pj T uj is the corresponding Weierstrass decomposition for T f . In fact, T pj is a distinguished j polynomial, by l), and T u is invertible. There remains to j fj

=

notice that a Weierstrass decomposition is unique. 4)

Let p be a distinguished polynomial. Then p

is

irreducible in zn (resp., en) if and only if it is irreducible n-1 [zn] (resp., cn' = @"-I rznl ) * in Zn'- 5 n' In fact, assume that p is reducible in 5 (resp., ? I ) . Then we may write p = f'.f'', where f', f" 6 Jn' (resp., cn') are assumed to be non-invertible. Write

f'

=

a ( z 1' ...,Zn-l) + 0

... +

LOCAL PROPERTIES

We may assume ah

bk

=

=

1, because ahbk

remains to replace f', f" (f'bk)(ft'ah).

119

by

=

1, and then there

f'bk, f"ah

Since f', f"

since p

=

are then monic, they both are

distinguished polynomials, by 2 ) , above. Since they both must then they both have non-invertible have degrees > 0 in z n' n' constant terms in 5 (resp., P ' ) , and so both are noninvertible in that p

zn

(resp., @").

is reducible in 3"

Therefore, p

(resp.,

p

f'-f'', where f' and

=

.

are non-invertible

...,O,zn)

and

p(0,

f'(0,

...,O,zn) #

0

f"

E

, , . . ,O ,zn)

Since p(0

=

(resp., ?),

zn (resp., Cn) f ' (0, . . . ,O ,zn) f ' ' (0, .

and

...,O,zn)

f"(0,

# 0.

p'u', f"

=

Weierstrass decompositions. Hence, we get p that is, p.1

=

By the Weierstrass =

p'f"', as

=

f'f"

(p'p'')(u'u'').

of the Weierstrass decomposition, we get that p

1

shows

# 0 (p being distinguished), we conclude that

Preparation Theorem, we get f'

(p'plf)(u'u'');

f'ef''

~n).

n Conversely, let p be reducible in 5 so that

=

=

By uniqueness =

pip''

and

u'u''. Since f' is non-invertible, then the degree of p f in zn must be > 0. Therefore, p' is non-invertible in 5n' (resp., n' ) . Similarly, p" is non-invertible in 5n' =

(resp.,

n'

C

)

so p = p'p''

is reducible in

(resp.,

zn1

5) If f

=

pu E

zn

(resp.,

pn)

is the Weierstrass

decomposition of f, then f is irreducible in if p

is irreducible in

p'

(resp. ,

el).

zn

if and only

LOCAL PROPERTIES

120

In f a c t , since f

zn

irreducible i n

in

3"

(resp.

,

(resp.

,

Proof of Proposition 7 : For

n = 0,

domain. n

-

(resp.,

3

@") i s e q u i v a l e n t t o

(resp.,

Cn) , and by 4 ) ,

3;"'

irreducible i n

n

i s invertible i n

u

Cn),

then

irreducible

p

t h i s i s equivalent t o

p

@"I).

The proof w i l l be by i n d u c t i o n on 0

we know t h a t

3

Assume then t h a t

n

C0

=

n.


= CC

1 and t h e proposition i s t r u e f o r

1. We a r e going t o show t h a t

f E

a;"

(resp.,

Cn),

# 0,

f

may be w r i t t e n as a f i n i t e product of i r r e d u c i b l e f a c t o r s .

Since

t h i s a s s e r t e d f a c t i s p r e s e w e d under i n v e r t i b l e l i n e a r change of v a r i a b l e s , we may as w e l l assume t h a t

f

s a t i s f i e s some

c o n d i t i o n (C ); t h a t i s , f has a Weierstrass decomposition P f = pu, where p i s a d i s t i n g u i s h e d polynomial and u i s invertible.

n' p E 5

Since then

(resp.,

?I),

by t h e

induction assumption and by t h e Gauss Theorem (Proposition 6 ) , (resp.

Zn'

write n'

n'

,

i s a unique f a c t o r i z a t i o n domain, and we may where

p = pl.. .ps,

is irreducible i n

'j

) . We may assume t h a t each

i s invertible i n replace

pj

by

n-1

(resp.

5

3;"

(resp.

,

= 1;

al...a @n-1 ),

a

,

j

hence each

a

S

j

and i t s u f f i c e s t o

a1 pj . Then,

j and being i r r e d u c i b l e i n

in

,

(resp.

i s monic, because i f

'j

i s i t s leading c o e f f i c i e n t , then

n;n'

c"),

by 2 ) above , each p i s distinguished, j n' 3 ( r e s p . , @ " I ), i t i s i r r e d u c i b l e

by 4 ) above.

Therefore,

f

has a f i n i t e

decomposition i n t o i r r e d u c i b l e f a c t o r s .

To showthat any

tux,

such decompositions

p = f l . . .fh

=

LOCAL PROPERTIES

gl.. .gk in

3;"

a r e e s s e n t i a l l y e q u a l , where

F fi(O,.

fi,

@"), we n o t i c e t h a t

(resp.,

..,O,zn) =

gj(O,.

121

are irreducible

p(0,

. .,O,zn).

...,0,z n )

=

Thus, each

. . ,O,zn)

fi(O,.

...,

...,

O,zn) # 0 , because p(0, O,zn) # 0 . By t h e j Weierstrass Preparation Theorem, consider t h e corresponding

g (0,

Weierstrass decompositions (p,.

. .ph) ( u l . .

.%)

=

f . = p.u i ' 1

(ql.. .qk)(vl.

of the Weierstrass decomposition, ul...%

- qjvj.

gj

. .vk) ,

p = pal =

and by the uniqueness

p = pl...ph

- ql...qk,

By 5 ) above, we s e e t h a t each

= vl...vk.

so,

1 = is

p. ,q 1 - j

n'

irreducible i n

3

(resp.,

Since

? I ) .

zn'

(resp.,

@"I)

i s by assumption a unique f a c t o r i z a t i o n domain, we g e t t h a t h = k in

n' 5

fjN qj

and a l s o , up t o t h e o r d e r of i n d i c e s , we have (resp., in

n

@"I),

(resp.,

;F

Eence i n

@").Hence

unique f a c t o r i z a t i o n domain. Comment:

zn

Q .E .D

(resp.,

sn

PjM q j @"). Therefore

(resp.,

en) i s a

.

Although t h i s proof by i n d u c t i o n on

n

is

n a t u r a l , we must add t h a t r e c e n t l y R a m i s proved t h a t t h e r i n g of convergent power s e r i e s on a complex Banach space i s a l s o a unique f a c t o r i z a t i o n domaim;

b u t the proof of t h i s more g e n e r a l r e s u l t

cannot proceed by induction on t h e dimension.


t h a t r i n g i s noetherian only i f t h e dimension of t h e Banach space i s f i n i t e .

# 0,

Bibliography

The following l i s t c o n s i s t s only of textbooks where the reader may f i n d f u r t h e r r e f e r e n c e s . Abhyankar , Local Analytic Geometry, Academic P r e s s . Behnke and Thullen, Theorie d e r Funktionen Mehrerer Komplexer, Veranderlichen , Springer. Bochner and Martin, Several Complex Variables , Princeton University P r e s s . Cartan, Elementary Theory of Analytic Functions of One o r Several Complex Variables , Addison Wesley. Fuks , Theory of Analytic Functions of Several Complex Variables , American Mathematical S o c i e t y . Fuks, Special Chapters i n the Theory of Analytic Functions of Several Complex Variables , American Mathematical S o c i e t y . Gunning and b s s i , Analytic Functions of Several Complex Variables , P r e n t i c e H a l l . Hervd , Several Complex Variables , Local Theory , Oxford University P r e s s . Hormander , An I n t r o d u c t i o n t o Complex Analysis i n Several Variables , Van Nostrand. Osgood, Lehrbuch d e r Funktionen Theorie (2 volumes)

,

Teubner

Vladimirov, Methods o f t h e Theory of Functions o f Several Complex V a r i a b l e s , MIT P r e s s . Wightman , "Analytic Functions of Several Complex Variables" i n Dewitt and Omnes, Relations de Dispersion e t P a r t i c u l e s Eldmentaires , Ecole d'e't6 de Physique Theorique, Les Houces , Universitd de Grenoble , 1960 , Wiley

.

122

,

.

Holomorphic functions, domains of holomorphy and local properties

Holomorphic functions, domains of holomorphy and local properties

Complex Analysis in Banach Spaces: Holomorphic Functions in Domains of Holomorphy in Finite and Infinite Dimensions

Extension of holomorphic functions

Holomorphic functions of several variables

Model Theory of Holomorphic Functions

Differentiable Functions on Bad Domains

Differentiable Functions on Bad Domains

Differentiable Functions on Bad Domains

Differentiable functions on bad domains

Differentiable functions on bad domains

Holomorphic Functions and Moduli I (Proc. workshop)

Generalized Functions, Properties and Operations

Measure theory and fine properties of functions

Measure Theory and Fine Properties of Functions

Singular Integrals and Differentiability Properties of Functions

Singular integrals and differentiability properties of functions

Measure Theory and Fine Properties of Functions

Measure theory and fine properties of functions

Differential Forms Orthogonal to Holomorphic Functions or Forms, and Their Properties (Translations of Mathematical Monographs)

Differential Forms Orthogonal to Holomorphic Functions or Forms, and Their Properties (Translations of Mathematical Monographs)

From Holomorphic Functions to Complex Manifolds

From Holomorphic Functions to Complex Manifolds

Holomorphic operator functions of one variable and applications

Topological Properties of Spaces of Continuous Functions

Differential calculus and holomorphy

From holomorphic functions to complex manifolds

Introduction to Holomorphic Functions of Several Variables, Volume II: Local Theory

Differential forms orthogonal to holomorphic functions or forms, and their properties

Nucleic Acids: Structures, Properties, and Functions

Differential calculus and holomorphy

Holomorphic functions, domains of holomorphy and local properties