Introduction to the theory of analytic spaces 1st edition (Lecture notes in mathematics, Vol 25)

Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, ZUrich Series: Forschungsinstitut for Mathematik, ETH, ZUrich 9Adviser: K. Chandrasekharan

25 Raghavan Narasimhan Tata Institute of Fundamental Research, Bombay Forschungsinstitut fur Mathematik, ETH, ZLirich

Introduction to the Theory

of Analytic Spaces 1966

Springer-Verlag. Berlin Heidelberg- New York

All rights, especially that of translation into fore/gn languages, reserved. It is also forbidden to reproduce this book, either whole or in part, by photomechanical means (photostat, microfilm and/or microcard)or by other procedure without written permission from Springer Verlag. 9 by Springer-Verlag Berlin 9 Heidelberg 1966. Library of Congress Catalog Card Number 66-29803. Printed in Germany. Title No.734~

CONTENTS

Page 1

Preface

2

Chapter

I.

Preliminaries

Chapter

II.

The Weierstrass

Chapter

III.

Local

Chapter

IV.

Coherence

Chapter

V.

Real

Chapter

VI.

The normalization

Chapter

VII.

Holomorphic

Bibliographical References

preparation

properties

sets

9 31 64

theorems

analytic

Notes

of analytic

theorem

91

sets

II0

theorem

mappings

of complex

spaces

123 134 140

PREFACE

The aim of these notes theorems

is to give proofs of the basic

in the local theory of analytic

of their applications of complex analytic holomorphic

The classical theory of n n functions in C and R specific r e f e r e n c e s

that are assumed.

algebra,

and factorial rings,

structure

spaces.

is assumed;

for the results commutative

to results on the global

and analytic

respectively

spaces and a few

are given

The elements of

especially properties and of elements

of N o e t h e r i a n

integral over a ring,

are also assumed. The term

"analytic

space" does not occur

In c o n f o r m i t y w i t h German practice, spaces over the complex numbers Further,

analytic

analytic

we have called analytic

simply

"complex

spaces over the real numbers

i n t r o d u c e d in all generality.

in the text.

spaces" are not

We have c o n s i d e r e d only

subsets of a real analytic manifold,

s a t i s f a c t o r y treatment of the general

since a

case w o u l d

involve

results of Cartan and Bruhat - W h i t n e y which we have but not proved.

Nevertheless

it was

title was the most appropriate.

stated

felt that the present

CHAPTER

I. -

The

aim

theorems

of

variables. are

of

shall

(Xl,

...,

a

Da

=

=

of

and

"'''

E

the

,

to be

following

f is

(

a! 0

f =

a =

x =

of

the

in H e r v 6

(a 1,

integers,

if is

= all..,

~j

results

[19],

If ...,

a n)

we write

(fl . . . . .

{ aj,

~ aj

of

fm )

(x 1,

...,

x n) ~C

lal

9' (a- ~) :

a function

a subset

an!,

for on

D,

is

(~)

we

n

,

= O

an

open

...

if

...,

We

n.

set

D

(in

from

~

to

Cn

sup

J

Ixjl,

,Xll =

or

R n)

If(x) I.

C

m

Ifi(x) I.

set 1

Ixl=

+ an ,

set

a mapping

we

= a1 +

otherwise.

j = 1,

llfll E = s u p m a x x~E i

If

various

complex

as m o s t

notations.

and

llfll E = s u p x~E

If

together

several

found

~an

~j

~n )'

~ { a If

of

proofs

non-negative

xn .

write

collect

an ...

(~1'

are

to

theory

give

X n ) ~ c n ( R n)

al = xI

is

[14].

use

an n - t u p l e

x

not

Proofs

- Rossi

We

chapter

elementary

shall

classical.

x =

this

the We

Gunning

is

PRELIMINARIES

(Ix1 12 +

...

+

IXn 12)-~

,

we

set

Let (real)

~ be an open set in

valued

holomor~hic

function

f

(or

defined

(real analytic)

is a n e i g h b o u r h o o d

Cn

Ur

Rn). A c om pl ex

on

~

is ca ll ed

if to every point and a power

Pa : Pa (x) = [ c

(x-

a~,

there

series

a) ~

s

which converges Proposition ~r n on

to

I.

f

on

A complex valued

is h o l o m o r p h i c

n

and,

the pa r t i a l

U.

for any

if and on ly if

j = I,

..., n,

~__~f = I ( ~f

~--" 0f

~'3

0~j)

Remark.

A basic

condition

that

f

shall not p r o v e

shall f.

f

on an open

set

is c o n t i n u o u s

if

xj = ~j + i~j,

derivatives

2 ,@~j

exist and are equal to zero in

If

fu nc ti on

t h e o r e m of H a r t o g s be c o n t i n u o u s

this here;

f = (fl . . . . .

say that

f

~.

f: m)

in

asserts ~

that the

is superfluous.

We

see eg.[18]. is a mapping,

~ ~ ~(Rm)

is h o l o m o r p h i c

(real analytic)

we

if the

are.

1

Proposition

2.

Let

the closure

of the p o l y c y l i n d e r

P = Prl,...,r n

Then,

f (x) -

for any

I (2~i) n

f

be h o l o m o r p h i c

[x~cnllxj

xs

in a n e i g h b o u r h o o d

ajlO,

of

l~<j, O

and e l e m e n t s

I ~ j { n,

such

that

n

=

[

j =i

~j

gij'

t h e n we h a v e

[cy

i = I .... m. y~

as an R - m o d u l e .

f

u.

= ~j,

the m o n o m i a l s

feR';

so is

so is

R' /

= k { x I ..... Xn},

= k { y I ..... ym} ,

q Yi

then

are of the

R = k{x}

1 { i { m,

fl .... ' f p ) ~

is q u a s i - f i n i t e , R,R'

is an i n t e g e r

gij~R',

(~I .... '~n'

is finite,

that

homomorphism, There

/

is q u a s i - f i n i t e ,

suppose

Let, R'

if

k{y}

[

with

a. < q, 1

i = l,...,m,

12

Clearly,

this can be written m

C2)

CaY ~i ~1c2 q [ ,~,

(if

c 2 > A2,

this last

series is convergent). Thus, we have (8).

Corollary. rings.

Let

an R-module

if

Let

c I ) A1, c 2 > A 2

u 9 R ~ R'

and c 3

be a h o m o m o r p h i s m

fl ...., pf ~R'. Then

fl .... , pf

if and only if their images

generate

the k - v e c t o r

Proof.

Let

space

satisfies

of analytic

generate f. 1

(14) ,

in

R'

as

R'/~R)R'

R'/~(R) R' D

R'/~(R)R'.

f.~R', 1 Then

u

1 ~ i ,< p

'

and suppose that

is quasi-finite,

f. 1

generate

so that, by Theorem 1,

15

u

is finite.

Let

R''

be the R - s u b m o d u l e

of

R'

generated

i

by the there

fi" is

R'/R''

By our

g~R''

such

= ~(R). (u

that

= R'

Theorem

being

2.

that

(R'/R''),

R-module, R''

assumption

R'/R''

finite),

that

p ~ O

and

h

f~R',

consequently

Nakayama's

g(x n) = f(O ..... O , X n ) ~ k { X n }

where

for any

is a f i n i t e l y

preparation

and suppose

g(x n)

fi'

f - g~R)R';

(Weierstrass'

f~k{x I ..... Xn }

on the

lemma

generated

implies

theorem.)

Let

the_element

i s not

identically

zero;

then

= x pn . h ( x n)

is a unit

in

k(Xn}.

We have

the

following.

(1)

For

and

any

~ k ( x I ..... Xn} , there

b I ..... b p E k { x I ..... Xn_l}

~

= a.f + (2)

There

exists

Furthermore in

(2)

i

(I)

and

u : R ~ R'

injection

a

Let

+

and the

are u n i q u e l y

Proof. let

b x p-I) u

n

such that

f = u.(xP

a~k{x I ..... Xn}

that

u E k { x I ..... Xn} w h i c h

a i .... ,ap~k{x I .... ,Xn_l}

aV

such

exist

is a unit,

au(O)

and

and

= O,

~ au.xP-U).n u=l b

in

(I),

and

u

and

determined.

R = k{x i ..... Xn_l},

R' = k ( x i ..... Xn}/(f),

be the c o m p o s i t e

of the natural

{x I ..... Xn_l} ~ k { x I ..... Xn}

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

k(x i ..... X n } ~ R' The a s s e r t i o n I, Xn'''''xP-ln to T h e o r e m

(I)

generate

I, we have

means R'

simply over

o n l y to prove

that

R.

the

images

of

By the c o r o l l a r y

that

their

images

16

in

R'/,~(R) R,

Now

generate

this

quotient

f(x I ..... x n) - f(O ..... O , x n) a~(R)R'

R ' / ~ ( R ) R'

space.

Thus we obtain

= k { x I .... ,Xn} / ( f , x i ..... Xn_l) = k { x I ..... X n} / [ g ( X n ) , = k(Xn}/(g(xn))

and clearly,

the

(2)

images

Apply

u ~ k { x i ..... Xn }

set

of

1,

(I)

.

.

= x

.

Xn,

' x pn - I

...

~ = X p. n au~k{x 1,...,xn_l}

and

x_ I

x I ..... Xn_ 1)

= k{Xn}/(xnP),

to

x p = u.f n

If w e

as a k - v e c t o r

-

There such

k{Xn}~xP) 9

exist that

~ a .x p-u. 1) n u---1

= 0

n-I

generate

in t h i s

equation

we

#

obtain

P x np = u ( O ' . " . 'O , X n ) g ( x n) since that

P = X n "h(Xn)

g ( x n) a

v

(0) = 0

This

and

proves

a

(O)

= O

will

u(O)

(I)

A polynomial

where =

and

a u (O)x pn - ~ ;

h(O)

1

/ O,

this

implies

~ O.

h(O)

(2).

P = xp + n be c a l l e d

[

a

.... , X n _ l ) X ~ -~ u (x i

a distinguished

(pseudo)

with polynomial

D

with

respect

to

To p r o v e first

that

functions

x n-

the u n i q u e n e s s ,

it is s u f f i c i e n t in

determined

we proceed to p r o v e

(I) ; in fact,

uniquely

u

by means

x p = u-l.f n To p r o v e ship

of the

uniqueness

-

in

the u n i q u e n e s s

and the

of the ~ = u=l

au

in

We

remark

o f the

(2) are

then

relation a x P-~ u n

(I) ; w e

remark

form P = af +

as f o l l o w s .

[ b .x p - ~ u n u=l

that

a relation-

17

in

R { x I .... , X n }

C { x I .... , X n } Hence

it

a

(the

suffices

Given exist

implies

v

f

as

the

in

of

relationship

in

x. may now assume complex 3 to prove the uniqueness when Theorem

~ C { x I .... , X n _ l } , n

neighbourhood

same

OeC

,

a

I,

(2)

implies

(O)

= 0

such

v

f(x)

--- 0

if

and

values) 9 k

= C.

that

that,

only

there

in

a

if

P xp + n

~ v=l

Hence

a x p- v = 0. v n

(continuity > O,

of

exist

6

E

fixed

x 1, 9 "''Xn-1

> O

f (xl , ....

zeros

Ix

now

that

we

where j

a,

r n-l,

b Ix

v n

I x 3I

' I < 6.

b'

a polynomial), small)

< e

of

(counted

+ = ~ b xP-U v n I

, a'

of

(arbitrarily

X n - 1 ' Xn)

l < 6 n have

~ ' af

roots

with

function in

the

,

j ~ nhas

xn

with

such

there that

I,

for

the

exactly

p

multiplicity).

Suppose

= a f + [ p '- vx b ~ n 1

are holomorphic v Then we have

Vx, I

in


~ O, 1 2~i

"

on

P'.

t k Of dt -~(x',t) f(x,,t ), Itl=r

n

If

tjl(x ')...tj(x~ ') [ l~Jl p}

in a

i.e.

~ck n , O~S.

set that

I = I (SO)

Proposition

4.

There

U = U'~U",

U ' c k p,

is a f u n d a m e n t a l

SnU

if

and

that P r o p o s i t i o n

to

p

if and o n l y

~

the r e s t r i c t i o n

of

a, Uc~,

such that,

set in an o p e n

of

S

independent.

is irreducible,

U " c k n-p

set

if it is not regular.

of d i m e n s i o n

= O,

of

= ~ . Choose coordinates n n,O 3 is satisfied. T h e n we h a v e

denotes

ideal

singular

fp+l''''' fn ~ n , a

be an a n a l y t i c that

a regular

a, S = {x I fi(x)

(dfp+l) a, .. -, (dfn) a

set in an o p e n

submanifold

is c a l l e d

is r e g u l a r

of

v

is a n e i g h b o u r h o o d

functions

neighbourhood

Let

is c a l l e d

is an a n a l y t i c

A point

exist

be an a q a l y t i c

if there

A point there

S

We h a v e

that a O(x') ~ O. m-1 + [ a (x')f u-I

0

kn

in

is so

s y s t e m of n e i g h b o u r h o o d s

such that

if

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

~ : (SnU) ~ U' of

U o n t o U',

37

then of

~ ~

is a p r o p e r is a f i n i t e

Proof.

Choose

polynomials

V = {xllxl ~

U"

a neighbourhood

< Q~ .

on

Since

such that

Ix

I ~ 2/2. r = {x"~kn-PIlx"l

fibre

V

of

~

-I

0

of P r o p o s i t i o n

analytic

> O,

then

and e v e r y

(x'),

x'~U',

set.

Pr(Xr,X')

coefficients

is

map

V

the

if

~ Q),

and

on

the

SnV.

Let

are distinguished,

Pr

if

that

3 all h a v e

and vanish

Ix' I ~ ~

Clearly

such

and

~ere

Pr(Xr,X')

U'

= (x'EkPllx'l

~

is as above,

= O,

< o}, then

- I (E) cEX {x" I Ix" I ~ Q/2}, so t h a t ~ is p r o p e r . F u r t h e r , -i if x~ (x') , t h e n P r ( X r , X ' ) = O, and Xr can take at m o s t Lemma

finitely

4.

Let

of analytic a finite

I

system

of

exists

O

such

induced

0

be

that

in

these

of

functions S

and

--oS the g e r m

set of c o m m o n

I.

system

set

~n,

as the

in P r o p o s i t i o n

a fundamental

Let

Pr'

r

3 and Lemma

zeros

> p, 2. T h e n

of n e i g h b o u r h o o d s are

in

analytic

U,

and

on

such

If

x~U'

8(x')

~ O}

are

= {x~Ul6(x')

= O,

• k n-p

~ O,

Choose

considered

U = U' U, S_o

that

(fi we

x U"

is

the

r

and x~U.

V = V'

x V" on

functions

generators find that

of

there

on

V

I). As exist

= O,

> p + !}. Pp+l(Xp+l,x')

then

analytic

/ O,Pp+l(Xp+l,x')

such V;

that

further 9

= O = 6x r - Q r ( X p + i ) ,

all let

the

functions

f1'''"

f 9

analytic

of

hold.

6x r - Q r ( X p + l )

Proof.

defined

as

values.

ideal

of g e n e r a t o r s

SqUn{xl6(x')

(b)

a prime

b y an a n a l y t i c

following (a)

different

be

set at

6x r - Q r ( X p + i ) there

many

with

SnV = {x~Vifi(x)

in the p r o o f f~,l" ~ p '

a =

be m

= O , i = I ..... m}

of P r o p o s i t i o n

2,

(ap+ I ..... a n), ~'3 < qj'

38

with f

i

and hence,

=

aj p + I}.

4. in P r o p o s i t i o n

4 and L e m m a

(which is a n e i g h b o u r h o o d

O~U',

this

is

that

a neighbourhood

O.

for a fixed

= {x~U 16(x') ~ O, Pp+I (Xp+l,x') This proves

This

= 6x r - Q r = O '

> p + I,

x V"

snun{xi6(x') and,

x k n-p.

then any s o l u t i o n

(2) i m p l y that,

If we n o w c h o o s e

V'

is d i s t i n g u i s h e d ,

r

= 0 = 8x r - Qr'

6Mf i = O ( m o d

on

Pp+I (Xp+l,x')

Since

neighbourhood

(I) and

- Qr )

V'

of

0

in

the a s s e r t i o n s

of

kp

0),

any

then we can find

such that

in P r o p o s i t i o n

4, g i v e n

for any open

4 and L e m m a

set

4

are true.

Proposition

5.

Let

U

be a n e i g h b o u r h o o d

L e m m a 4 is true r e l a t i v e is i r r e d u c i b l e .

Then

a regular

of

has

point

a jacobian

to the

any p o i n t S

ideal

I = I(So)

x~SnU

with

of d i m e n s i o n

of r a n k

p

at

of

x.

p,

0

6(x')

such that where ~ O

S --o is

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

40

Proof.

Since

6(x')

that,

at the p o i n t

near

x

b y the

Pp+i(Xp+l,x')

~ 0

x,

Pp+i (x) = O, we c o n c l u d e

and

@Pp+i / O. ~Xp+ i

system

Hence,

S

of e q u a t i o n s

= O, x r =

~r (xp+i) 6(x')

' r

that

k-independent

at

Remark countable

d P p + l ..... d(x r x.

further

at

~

that

f-l(y)

dim X

{ dim Y

that

and

if

X,Y

of

with

5 we o b t a i n

Proposition relative largest set S

6.

integer

integer

m

This c h a r a c t e r i s a t i o n invariant O

in

at

analytic

3.

0

this

2 and

3

is the

is i n d u c e d b y an a n a l y t i c contains

points

at w h i c h

m. of the i n t e g e r

automorphism

a~k n

The dimension is the i n t e g e r

of an a r b i t r a r y

analytic

of the i r r e d u c i b l e an a n a l y t i c where

Combining

p

is c l e a r l y

of a n e i g h b o u r h o o d

of

kn .

Definition ~a

under

rank).

irreducible,

of

such

to a p o i n t w h e r e

being (~a

map

then

of P r o p o s i t i o n

such that

of d i m e n s i o n

manifolds

p

every neighbourhood

is r e g u l a r

y~f(X),

theorem

has m a x i m a l

_ a) , _S a I = I(S

to

S),

f

The

are a n a l y t i c

for e v e r y

the d i f f e r e n t i a l

5.

is an a n a l y t i c

(apply the r a n k

Proposition

Proposition

f : X ~ Y

is d i s c r e t e

are

6 (x'))

This p r o v e s

w h i c h has

> p + i,

Q~ the p r o p e r t y

is d e f i n e d

~a

set

S

germ

components in

is the 9 e r m

an

open

of an i r r e d u c i b l e

analytic

p

2. The d i m e n s i o n

of P r o p o s i t i o n

_

Sa

is the m a x i m u m

9erm

dimension

of

S of S 9 The d i m e n s i o n of --v, a - --a is m a x d i m Sa, set ~ i..~n k n as

at a d e f i n e d

by

S.

41

Theorem

i.

Let

in

k n.

of

a

p.

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

--

Let

S

a~S

contains

dense

in

Let

and

T --a

be

Let

components

of

S a"

in an o p e n at

T

T" .

Let

to

T . --a

Then,

(relative

Proposition

any

regular

points

set

Q

neighbourhood of d i m e n s i o n

of

S

is

be

= -Ta nT' -a

is

Let

6

and

= TuT".

Since

does

not

point that

-S -a

other T,

be

inducing

the

a regular

it

T

of

irreducible

T"

lie on

p, of

of

analytic germs

point

T"

is c l e a r l y

suffices

to p r o v e

of d i m e n s i o n

p

Propositions3,

4,

not

5 apply

f~9

f = O on T" f~I = I(T a) n' --a' -h a v e the s i g n i f i c a n c e of P r o p o s i t i o n 6~I.

Then

a,

near

S

is

--a

S' --a

Since

there

i follows

where

the

a

so c h o s e n

If

of

of d i m e n s i o n

there

7.

union

which

S

component

are

from

an is

I

is p r i m e ,

points

x~T

Proposition

irreducible any

analytic

f6~I. with

5.

germ

of

germ

at

dimension a,

then

< d i m --a S . Clearly,

choose

I = I(S a) _

r

p

T ). --a

S ~S', --a --a

Proof 9

there

an o p e n

is r e g u l a r

containing

SnU'

Theorem

f (x) ~ O.

We

U

to

T"

-a

U'

arbitrarily

d i m --a S'

in

Then

S

the

a regular

T" (T') --a --a .

and

= p.

of

be

Let

of

contains

on

p

T'

then

point

U

Hence,

set

irreducible

--a

set

a;

a regular

(since

set

an

of d i m e n s i c n

that

S

--a

at w h i c h

the

p.

of

analytic

dim

points

dimension

T T" --a' --a

an

S.

Proof.

sets

be

the ,

I'

exists

> p.

If w e

we may

suppose

x i, 9 .. ,x n

coordinates = I(S a)

(so t h a t

a distinguished show

that,

_' Sa

that

in

IcI') ,

is

irreducible 9 so t h a t

kn then

we

a linear

have

Ip

=

Pr(Xr; x')~I

pseudopolynomial

after

if

change

of v a r i a b l e s

in

k P ( x I .... ,x ), t h e r e is a d i s t i n g u i s h e d polynomial P Pp(Xp;Xl, .. 9 I) ~I', the r e s u l t f o l l o w s f r o m the r e m a r k after

Proposition

suffices Let

2. Now,

to p r o v e

g ~ O n,

g~I,

that

g~I'

by

there

the

preparation

exists

By L e m m a

3,

h s p nI' there

is

theorem, = I' p' gls

it h ~ O.

-

I

{0},

42

such that since

ggl ~ h(mod

gr

that

h~I',

h~I,

analytic

In this

I

Proposition

so that

is prime,

= ~ , and suppose n are so chosen that

in ~ n , o

(x I ..... Xn)

defined

4 are valid. in Proposition

system of generators

independent

of the system of generators

Proposition

8.

We have

Since

Hence

it suffices

~

x'r

~(SnU)

/ O,

Xp+ 1. Let

x~SnU,

contains

the dense

Remarks.

I. Note that,

Proposition

that,

neighbourhood onto

of

O

in

map.

0

C p,

CP .

(x

in in

U' U'.

Pp+l(Xp+l,X')

has

Qp+2 (Xp+l) Q n (Xp+l) 6(x') ''''' 6(x') )"

= x'

Hence

~(SnU~

following

Proposition

are so chosen

is a fundamental

3,

that system

such that ~

being

~ (SnU) is a p the projection of

Cn

P 2. Actually,

- s

then there of

is

~ O).

by the remark

{Uu}

is dense

~(x)

if the coordinates

of neighbourhoods

is closed

x : (x',Xp+ 1,

{x'r

2 is valid,

S(I)

chosen.

the polynomial

and clearly set

Clearly,

= U'

to show that

4,

implies

4.

I.

its image

By Lemma

this

~(SnU)

of

is proper,

6(x')

zero

~ : SnU ~ U'

we denote by S(I) the germ n,a set defined as the set of common zeros

of a finite

Proof.

Let

Ic~

of analytic

a complex

so

h / O.

ideal

3, 4 and Lemma

For any ideal

For any

ggl~I,

k = C.

be a prime

be the projection

a

I

But then clearly,

sets.

that the coordinates

at

since

h~%.

section we shall deal only with complex

sets,

Let

and,

where

and in particular

w 2. Complex

analytic

I),

uI6(x')

0}

the map u'

-

16(x')

is a covering

43

Lemma

5.

above, that

Let

fE~

. n x'~U',

for a n y ~(x)

Proof. where

= x'

If h~

and

f~I, .

If,

6(x') f(x)

there

Then

for s u f f i c i e n t l y

h~I,

~ O,

= O,

is

hhere then

g~I

small

is

U

x~SnU

as such

f~I.

such that

gf - h(mod

and for a n y s u f f i c i e n t l y

I)

small

x,

P

6(x') / 0, h(x') so t h a t Theorem

h = O 2.

ideal of defined

= f(~) g(x) and so

= o

h~I,

(if

x~SnU,

a contradiction.

(Hilbert's Nullstellensatz).

~n

and

~(x) = x'),

--oS = S(%)

Let

05

the g e r m of a n a l y t i c

as the set of c o m m o n

z e r o s of a f i n i t e ~

generators some

integer

Proof. This

We

m

(i.e.

since

decomposition

that,

consequence

rad~

if

is prime),

that

~

is prime,

of Lemma

for

I(~o)

= rad~.

~

is n o e t h e r i a n , n theorem,

l(So)

5. Hence,

w e deduce,

k = u/~--i~ v ' Clearly

s y s t e m of m

{f~(Dnlf'"~ 0%

I(S O ) = rad0% =

first remark

= S(rad~),

{O},

Then

set

> 0}.

is a t r i v i a l

primary S(~)

of 0~ .

be a n y

If

if

=~. ~

is

since 05

is a r b i t r a r y ,

w e o b t a i n b y the N o e t h e r

~u b e i n g p r i m a r y .

then k

s(~) = ~ I

s(9),

so t h a t

We now give obtained

above.

k

k

u=l

v=l

a very important

if t h e r e

holomorphic

= rad~.

application

of the r e s u l t s

We begin with a definition.

D e f i n i t i o n 4. Let n .in C A function a~S

rad ~ u

S

be an a n a l y t i c

f

o_~n

S

F

in

set

is said to be h o l o m o r p h i c

is a n e i g h b o u r h o o d

function

set in an o p e n

U

U with

o_~f

a

FIUnS

i__nn Q

and a

= fIUnS.

at

44

We m a y d e f i n e obvious

way.

If

of h o l o m o r p h i c

germs

a~S,

of h o l o m o r p h i c

let

functions

Definition open

5.

set in

A map C ni)

denote S.

ring over

f : Si ~ S2

is c a l l e d

the ring Clearly,

/-/T(Sa) ~---~ -S,a

is an a n a l y t i c

S,a

S,a a on

at

n,a Hence

~

functions

in the of germs we have

D

C.

(S i

analytic

holomorphic

set in an

if the m a p

j o f,

n2 where

j : S2 ~ C

jof =

(fl .... Clearly,

Theorem

form

) w h e r e the f are h o l o m o r p h i c on v 2 a holomorphic map f : S I ~ S 2 induces,

an algebra

f*(9)

has the

Si

n

"

for

homomorphism

f* viz,

injection,

f

9

a~S 1,

is the n a t u r a l

: ~)S 2 9 f(a) ~ )

Sl,a

= 9 o f-

3.

Let

be h o l o m o r p h i c .

f : S1 ~ S2

Then

the

homomorphism

f* is finite point

: ~ S 2 , f(a)

(see C h a p t e r

of the

fibre

II)

if and o n l y

if

a

is an isolated

f-lf(a).

Proof.

Let

Slr162

We m a y

suppose

that

n) , a = O,

f(a)

RI = ~ S 1 , O " Suppose

~ ~ Sl,a

S2r162 = O.

R2 = ~ S

).

We set

2'

f(O) "

that

f*

is finite.

Then

every

:

element

R 2

of

~

RI Ri

is integral

over

R 2.

45

Hence,

if

e~S

al,...,ar~

%

2 '0

there exist h o l o m o r p h i c 1,0' such that

germs

r r

v

* [

o.

V=l

In particular, r [

r xk +

Hence,

a

~=1

if

satisfies

we have, (k)( v

f(x)

f(x)

in some n e i g h b o u r h o o d

) Do,

A~,a = A~o,a )' A~nA~o

of an open set

in ~ n , a

~n,a

be the irreducible

~o

S

K.

local, we may suppose

This would clearly conclude

since

A (~)

~,

subset

a maximal

by the maximality Let

for any compact

we may clearly suppose that

is a neighbourhood

we have,

Then,

a family filtered

the family is stationary on

Thetheorem X

{A~}

3.

to the family of finite

71

Proposition

4.

Let

s u b s e t of

X.

s u c h that,

for a n y

If t h e r e any

and

is a n a l y t i c Proof.

Let

X - Y

which S~.

B )A,

d i m --a A = p set

if

p o i n t s of

p

(on

some

~(a).

S~ ~ @, B~,

B~

B~ (a) "

B - Y,

Since

irreducible,

X),

is

of

B

that of

a

B~

and

X

into A

B" - Y,

is the set of r e g u l a r hence

A a = S~(a) .

p o i n t s of

A a r S~(a) = p

for

S~ (a)

and

Further,

S~(a)

is

is d e n s e

Thus =US

= U B~

be a c o m p l e x

s u b s e t of

X.

Let Let

such that In fact,

components

of

Proposition

Let

be a h o l o m o r p h i c

is a n a l y t i c

if and o n l y

U

space,

Y

s u b s e t of

if t h e r e

an a n a l y t i c X - Y.

is an a n a l y t i c

is t h e n the u n i o n of t h o s e are n o t c o n t a i n e d

X, Y

map and

f-lf(a).

neighbourhoods

be an a n a l y t i c

which

5.

at a n y p o i n t .

set

B - Y = A. A

B

X

A

is a n a l y t i c

p o i n t of

in

its p o i n t s ) .

is c o n n e c t e d ,

4'

Proof.

A

is the set o f

if

Proposition

X

of

for

then

p

in

= p

a.

is i r r e d u c i b l e

of dimension

B

of

B~ - Y = S~

is a n a l y t i c

A

dim ~ A

A = U A

A a r USa'

A = UA

Then

X - Y

independent

with

d i m A a = d i m S~(~)

we have

an a n a l y t i c

s u b s e t of

at a n y o f

that

is a g a i n c o n n e c t e d , Aar

Y

b e the d e c o m p o s i t i o n

We r e m a r k

since

Be X,

space,

then the c l o s u r e

components

Since

in

a{A,

B = U B~

X - Y).

regular

b e an a n a l y t i c

(of d i m e n s i o n

irreducible

in

A

be a complex

is an a n a l y t i c

b rB

(on

Let

X

Then

be c o m p l e x a~X a

be such that has

Y.

spaces, a

a fundamental

s u c h t h a t the g e r m of

of d i m e n s i o n

in

= d i m --a X .

irreducible

f(U)

let

f : X ~ Y

is an i s o l a t e d s y s t e m of at

b = f(a)

72

Proof.

Clearly,

set in an open open

C .

{(x,y) ~n• and

if

suppose

Chapter

p r o j e c t i o n of S into ~', -I point of ~ ~(Zo). It suffices

at

III,

z o.

Theorem 2, that ~ere

in

C k,

Let and

U = U'

x U",

to prove

III,

such that 2,

is dense

SnU.

in

restrictLon

to

U'

• U~ • ...

in

C (Xp+ h) ],

d imens ion

U

U'c

C p,

the

Then,

SnU

for

j ~ 1,

then

~.3 (S)

is a n a l y t i c

is

III,

of c o o r d i n a t e s of

0

U"r

C m-p,

in

C m, such the III.

statement.

x U",

set

With U'r

~. 3 of U

the

III,

denotes

~ O} the

onto

• U"n_p, in

in

C p,

Sn{xcU]6(x')

if

[U" = U 1'' • ...

S

from

of C h a p t e r

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

• U~3 = U.3

that

is irreducible.

such that

o

We m a y

set in an open

U = U'

= s

to an

satisfies

following

S --o let

and

to

4 of C h a p t e r

the c o n c l u s i o n s

3, 4 a p p l y

thus

of C h a p t e r

change

be an a n a l y t i c

as in C h a p t e r

f(a))

Q'

suppose

W,

(a,

It follows

4 and Lemma

that

be

onto

~ : S ~ (V • U) ~ U'

S

Propositions

o~

a linear

suppose

C n-p

we m a y

3, and the p r o o f after

in

is the r e s t r i c t i o n

dim S_o = p.

of P r o p o s i t i o n

O~S,

U.3

U~

open

of

p. (1.1) .

in C h a p t e r

of the

f

are n e i g h b o u r h o o d s

it is e n o u g h

notation

as

Let

the p r o j e c t i o n

Proof

when

Clearly,

(1.1)

U"r

is a n a l y t i c

z O = O.

conditions

C n,

S

that

0

Thus,

then

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

cm()~ ')

that

is an

is the set

Sc ~x~'

Proposition

of

is an a n a l y t i c Y = ~'

set

irreducible

V

y = f(x) }

X

and that

Sc W = DxQ'

the p r o p o s i t i o n

analytic

in

Ck

in

that

denotes

is an i s o l a t e d prove

R If

I x~X,

~

suppose

set m

set in

we m a y

III,

Let

P39 ( x P+J

Proposition

;x')

,

8x

3. Then

P+J

-

SnU

Qj (Xp+ i) ....

be

is the c l o s u r e

set

{x~Ul6(x')

~ O, Pl(Xp+l;x') , 6Xp+j

- Qj (Xp+l)

= O,

j = • .... ,n-p}.

73

Hence

is the c l o s u r e

~.(S)

of the

set

J A = {(xl,...,Xp+j)

~Ujl6(x')

6Xp+ 1 - Ql(Xp+i) But if then

B

is the

Bn{6(x')

# O, Pi(Xp+i;x')

= O,

set

1 = i .... ,j}.

{Pi(Xp+i;x')

# O} = A;

= 6Xp+ 1 - Ql(Xp+i)

by Proposition evident

~. (S) onto 3 Propositions

U'

the c o n d i t i o n s

2, 3, 4, so that

We sheaf

shall

of an open

without real

change

which

we

use

set

Of course, n to that of C

Theorem

3.

(K. Oka)

Let

= O

denote

the

holomorphic

functions.

of an open

of the sheaf

later.

reduced

~

sheaf

in an open

be t r i v i a l l y

let

III,

dim ~ [~j (S)] = p.

to the case

functions

shall

of C h a p t e r

set in

C

n

f o l l o w i n g Oka, that the s t r u c t u r e n C is coherent. The p r o o f applies

set in

also

analytic

that

of the s t r u c t u r e

now prove,

i ,< j},

A = ~. (S) 3 the p r o j e c t i o n of

It is further satisfies

= O,

4' above,

is analytic.

2. The c o h e r e n c e

= O,

~

Then

in

~

on

R ,

the case

be an open

sheaf

of germs n

~

set

a fact of

in

of qerms

Rn

can

n C ,

and

of

is a c o h e r e n t

n,~

of

sheaf

of

rinqs. Proof.

Since

it suffices

to prove

For any induced

by

Theorem

4.

and

~

let

for any

(~i ....

trivially

On, Q

the

~%,~,

is of finite

type

over

itself,

following.

(~) a

will

denote

the g e r m

at

a

e.

Let =~(fl

as ,aq) ~ q

O~ a

n a 9

fl .... 9 .... 'fq) is the with

be h o l o m o r p h i c denote

the

submodule

of

sheaf ~q n9

~ ai(f i) = O. a i=i

functions

on

of relations 9 consistinq Then

for any

Q9 i.e. of ar149

74

there exist s I .... ,s k for any

a neiqhbourhood of

~

b~U,

Remark.

i.e.

Theorem

and the direct

assertion

which generate

Consequently,

4 is proved,

Then,

4.

~Pn-l'n' Let

clearly

suppose

n

above,

f 'q

and

it follows

n,~' is

~,

the

that the sheaf of ~ n P ,~

is by induction

is coherent

on

for any open

for any sections

' ~ ( g l ' " .. ,gq)

fl ....

of

is of

p ~ 1.

The proof

by our remark

~b

sheaves

finitely m a n y sections

On -I,~'

that

over

to the coherence

if for some

type for any integer

Suppose

~b

is of finite type.

4 is equivalent

between

Proof of Theorem

of

~

in Theorem

of relations

U

and finitely m a n y sections

sum of finitely m a n y coherent

again coherent.

finite

over

U

is of finite

be holomorphic

on

n.

set

~'r

gl ..... gq

type over ~r C n.

~n-l,n'"

We m a y

f. ~ O for each i. Since Theorem 1 and permits m u l t i p l i c a t i o n by units, we can

is local,

that

assume,

after

a linear change

a = O,

and b y the p r e p a r a t i o n

of coordinates, theorem,

cn-I .

4

that the point

that

pi-1 9 =

fl

P.

9

=

xPi n

+

a(i)

~

(

xI

.....

Xn_ i

)x v

n

v=o

(i) where a (i) a (O) = O.

is a holomorphic We m a y suppose

that a relation polynomial,

that if

over

~b'

~. 1 holomorphic

~ = n' x D,

(2.1)

that

(~I ..... aq) ~ ( P l

if each

coefficients

function

For anv

(b',bn)~

b y the polynomial

For the proof of

b'~C n-l,

D = {Xn~CllXn~ b =

r

cn-l,

P = Pl = max Pi" i ..... pq) = ~ is

is a polynomial at

~,

on

b = < Q}

' ~b

relations

(2.1), we need

in

in

Or We say

x

with n (b',b n) . We claim then: is qenerated, xn

of deqree

p.

75

Lemma

I.

Let

Q(x';Xn)

= xqn +

[

be a

a~ (x')xq-vn

v>o

distinguished

pseudopolynomial

is d i v i s i b l e element

of

by

Q

in

at

O.

Then if

R~n_l,o[X is an

the q u o t i e n t

~n,o

n]

~n_l,o[Xn].

Proof

of L e m m a I. Let U = U' • D be a n e i g h b o u r h o o d of n implies x ~D. O in C such that x'~U', Q ( x ' ; x n) = O n Let R be Then, for any x ' ~ U ' , Q has q zeros in D. holomorphic

in

U,

RE%_l,o[Xn],

holomorphic

in

U.

Then any zero

a zero of

R

algebraic

U.

division

holomorphic ~r

in

,

Since

over

functions

deg

~ < q,

and

Q

of

in

U',

Q

be

U

in

is monic,

the ring

R/Q

let

is also

we m a y m a k e

A = F(U';%_I,U,)

so that

~ere

an of

exist

9,

with R = ~Q + ~.

For any in

x'~U',

D;

Q

hence,

and b e i n g

and

R

for fixed

a polynomial

have x',

at least

~

of degree

q

common

has

at least

< q,

~ ~ O.

q

roots

roots,

Hence

R/Q = ~ A [ X n ] Note. as one

Lemma

sees

.P.roof of

i is v a l i d

at once b y

(2.1).

where

Q

leading Lemma

I,

Clearly, (PI

b =

in

term),

= Pl(X';Xn)

~n-l,b' of d e g r e e

u~%_l,b,[Xn], for

in the

i > 1,

functions

= u . Q ( x ' ; x n - b n)

polynomial

(vanishing Q ,< p,

at

and

the e l e m e n t

we w r i t e

x n - b n' w i t h

in b' u

and has d e g r e e

i-th place)

(~l'''''~q) ~ b '

analytic

Write

(b',b n)

is a d i s t i n g u i s h e d

coefficients

for real

"complexification".

Let

p(x';Xn)

also

si =

except

for the

is a unit.

By

p - Q(4 p) . (-Pi,O,...,O,PI,O,...,O)

is a p o l y n o m i a l ~i = ci Q + ri'

relation. where

If c i , ~ n,b,

76

ri~@n_l,b,[Xn]

and

deg.

this can be w r i t t e n

r i < 2;

~i = diP1 + ri

d i = c .i u - l ~

'

n,b"

Hence

(a I .... ,~q) - d2s 2 -

where

-

d

#1(%,b , ri~%_l,b,[Xn].

qsq

=

(~1,r2

Clearly

.....

rq),

(#1,r2 .... ,rq) ~ ,

so that C

_~1P1 is

an e l e m e n t

of

b y Lemma 1,

=

~)

r.P. 1

1>1

~n_l,b,[Xn]

131ur

uri~%_l,b,[Xn]

-

of

degrees

and has

and has d eg re e

1

< p + 2-

degree

< p.

< p.

Hence,

Also

Thus -i

(~I ..... aq) and since

= d2s 2 + ... + d q s q + u

s 2 .... ,s , (~l,u .... ,r u) q q of degree { p, the a s s e r t i o n

relations

To c o m p l e t e only to prove (2.2)

(v)

the proof of T h e o r e m

There

exist

(X'l~V) ,---, (V))q

b~U

are p o l y n o m i a l (2.1)

is proved.

4, we have t h e r e f o r e

the following.

of the o r i g i n at

(~lu,r2u,...,rqU)

finitely many polynomial of deqree

~ p

in a n e i q h b o u r h o o d

such that any p o l y n o m i a l

is generated,

over

~b'

relations

relation (v) by the

U

of d e q r e e {

p

P Proof

of

(2.2).

Let

~ =

-(~1 . . . .

'

~q)

'

~.

i

=

c P( i ) ( x ' ) x

n

V =0

be any p o l y n o m i a l 9 = Pl

P [ P=O

relation

a (i) (x')x v v n"

at

(Note:

b = (b',bn), a (i) Pi - i,

and wr it e

(i)

aP

= 0

if

u > pi )

77

Then,

~L is a relation if and only if

q [ ~ i=l k+ = v

~ i ) (x,)cl(i) (x,) = O

in ~ n _ i , b''

for each

v;

(p+i) q (c(1) o '" .. 'c p(q)) ~@n-l,b'

this clearly means that the element is a relation between the sections

(i) ( x ' ) ( ( i ) (x') ... a (i) (x')) Sk = ao-k ' ' p-k O { k { p,

where

a (i) = O

if

of ~ p + l ~'n-l,U''

~ < O.

I~ i ~q

The statement

(2.2)

V

is now an immediate consequence of our induction hypothesis. We note explicitly the following Corollary.

D

Let

be two subsheaves of ~ , ~ _ is also of finite type

finite type. Then

of

(and so

coherent).

Jl

In fact . if. .(f~l){ .. f(~)) (fl, 2)( .. , f(2)) generate _ . q ' _ ," q ~ at every point of an open set U respectively, then

~i Fl~2z is a quotient of

~=~(f~l)

,---,fq(') , f(2) ,--.,fq(2)) :

it is the image under the homomorphism ~ - ~ , ~ , . ql sends

(~I ..... ~ql + q2 )

to

which

e=l[ av f(1)v

3. The coherence of the ideal sheaf of an analytic set Theorem 5. Let

Let

~ = ~(S)

for any

a~,

S

be an analytic set in an open set

be the subsheaf of

f~

a

belonqs

vanishes on the qerm

-

S

--a

to

~

defined as follows:

O~a i f and o n l y i f

induced by

~( C

S

at

m

a.

f

Then the

A

sheaf Remark.

J

is coherent. The corresponding theorem for real analytic sets

is false, as we shall see later.

n

78

Proof.

Since

~ r

to prove

that

~

3a = ~a;

hence

on

fl - S.

of

a,

2 ~

S I .... ,S k

components

of

each

~(Sv)

prove

that

aES,

and

Sa .

U

sets

is

of

P = dim --a S ,

then

III,

U = U'

is a d i s t i n g u i s h e d

U

(S~) ,

of

a neiqhbourhood

of

suppose

O

such

Proposition U'r

P

(x

coefficients

and p o l y n o m i a l s

Qr(Xp+l;x')

of d e g r e e

if

6

is the d i s c r i m i n a n t

of

PI'

that

n

the If

r ) I;

;x') p+r on U',

of Pr~3(S)

< k = kI

then

a = O~C

satisfied.

for any

with

k

that

3 are

Cp

polynomial

to show that to

in

We m a y

so that , b y

it is s u f f i c i e n t

r holomorphic

degree

type

such that

only

type

x U",

then c l e a r l y

are the i r r e d u c i b l e

Thus,

be a n e i g h b o u r h o o d

a~S,

4,

be a n e i g h b o u r h o o d

in

4, we have

is irreducible.

of C h a p t e r

U

U = ~

finite

Theorem

and so is of finite

S

type.

after

If

let

We have

to T h e o r e m

Sa

Q - S

analytic

~(S)

conditions

on

is of finite

when

there

= ~

type.

and the germs _

the c o r o l l a r y

Let

is of finite

Let n o w

SnU = U (S nU),

a~S

it is sufficient,

suchw that

6Xk+ r - Q r ~ ( S )

and (a)

Sn{x,Ul6(x')

~ O} = {x,Ui6(x')

(b)

Sn{x~U 16(x')

~ O}

We c l a i m

the

(3.1)

is dense

/ O, Pl(Xp+l;x')

in

following. There

set,

is an inteqer

for any

V,

and

b~V,

by

~I ..... ~ i"

Proof

of

Then,

since

Qr=O}

SnU.

N

> O

~(Sb)-- = ~b = { f ~ % I 6 N f ~ I b ( P l ' ' ' ' ' P n - p ; we have

= 6.Xk+ r -

i ~I .... ' @I

Let

Pr'6X

for any

- Q2'''''6Xn on an open

generated

in

b~U - Qn-p ) };

set

~b

0~b = { f , @ b l 6 N f , I b ( P r , 6Xp+ r - Qr ) }. - Qr

p+r

6Xp+2

holomorphic

Ib(~l ..... ~i ) = ideal

(3.1).

such that

vanish

on

SnU,

6Nf~b

if

f~b"

79

But since the set

Sn{xeUl6(x')

this implies that

fe~b'

Let near

so that

b = (b',b")eU.

Xp+ r = bp+ r,

Pr = UrAr,

~ O}

Then clearly

functions of < i

p+r

~ O

Ar = Ar(Xp+r - bp+r;X') ~ p , b ' [ x p + r ] ir

and

is

u r e~n,b

is

denotes the ring of germs of holomorphic xl, ...,x

at P ~ae%,b''

we can find germs a

Pr(Xp+r;b')

so that, by the preparation theorem,

where

(~p,x'

SnU,

~ b r ~b"

a distinguished polynomial of degree a unit

is dense in

x' eC p) .

Hence, for any

~b'

a = (ap+ I ..... ~n ) ,

so that

r

(x) - a

ap+i an ~a (x'")xp+l "''Xn ~Ib(Al ..... An-p)

~ I,

so t h a t

= 0

D I x Wo

Further, imply

if

that

can = 0

x

is P

find neighbourhoods on

we

v I,

WO

Since

v2,.--,v r 9

we

w(x,w)

near

on

w O.

near

points,

is h o l o m o r p h i c

the p r e p a r a t i o n

of

- Xl;W)

is n o n - z e r o

near

- Xl;W)

VinZ.

can write,

by

theorem, q-I

-I

~w

=h

[

Q +

d

(w) x ~,

q = deg

Q,

v =o

h~Ov,Vl 'dr(w)

where

is of d e g r e e

'~W,

p - q,

Wo"

Hence,

we have,

since

and

~OW,wo[X]

in a n e i g h b o u r h o o d

of

V1

vI,

p-1 = h P +

q-1

p-i

[ c cw x~ =

where

[

v=o V. nZ, 1

i

~ c (w) x v, v v =o

0cw x }

Since

(4.2),

so t h a t

w

= 0

on

v=o

> i,

this

gives

us

(4. I.A)

is

proved. (4.I.B) Z with p-I c v=o

u ( v i)

then

of

Wo~W,

= Wo,

(w)x v = O

v

U {vi}, Proof

If.

on c

and

and

if

ZnN

=

(4.I.B).

O,

v l,...,vr c ~% v ,w

where

v

Let

=

0 .....

~(w)

polynomial

P(x;w).

Since

P

the o r i g i n ,

the germ

of

at

A = fw,wl

Cw)

=o).

w-

A

N

be

'"are t h e

points

and o is a n e i q h b o u r h o o d

p

-

h a s no m u l t i p l e

is d e n s e

of

I.

the d i s c r i m i n a n t

w = 0

of

is n o n in

W.

o f the

factors zero.

Let

at

87

Let

w be a p o i n t of W - A near (p) be the d i s t i n c t roots of

(I) X

, ...,X

(x

and if

w

is n e a r e n o u g h

w o,

and

P(x;w) to

let

= O.

Wo,

Then

(x (i) ,w)~N.

p-i Hence,

cv(w)xV

has

p

distinct

roots

if

wo

is near

V =O

w,

w~A.

Hence

for all such

w,

we h a v e

c

(w) = O, V

0 { v ~ p - i. implies

Since

that

c

W - A

= 0

is d e n s e

in

W,

this

in W,w O "

The

following

Proposition

corollary

5), but we

Corollary.

If

an a n a l y t i c Proof. corollary

map with

spaces

finite

from T h e o r e m

and

fibres,

(in 7.

f : X ~ Y

is

f (X)

then

is

Y.

clearly

follows

it follows

are c o m p l e x

subset of

Since,

already been proved

show that

X, Y

a proper holomorphic

has

f(x)

at once

= {y,Y I f * ( ~ X )y ~ 0},

from T h e o r e m

7 and the

this

following

proposition.

Proposition analytic stalk

6.

Let

sheaf on

~

of

X

X.

~

be a c o m p l e x

space,

Then the set of

at

x

is not

the

x~X

0

~

a coherent

for w h i c h

module

the

is an a n a l y t i c

X

subset of

X. The p r o p o s i t i o n

Proof.

is an a n a l y t i c

is ~

set in an open

to

~

~

- coherent.

~ ~

exists

by

~ ~ if

0

0 ~

being

outside

local, set

X,

~

we m a y s u p p o s e that X n If we e x t e n d in C

the sheaf

~'

H e n c e we m a y s u p p o s e

is an exact

is small

is g i v e n b y the m a t r i x

sequence

enough), A(x)

thus o b t a i n e d X = ~.

that

on

n

If

(such a s e q u e n c e

and the h o m o m o r p h i s m

= laij (x))I ~ i { p

A

: ~P

of h o l o m o r p h i c

I~ j~ s o.

III,

S ) (A - A') nU.

{As}

the p r o p e r t y

> s o.

of d i m e n s i o n

X~

of

property

Iao

s

V

and

is m a x i m a l .

having

of

and hence

a~

Iao

A'r A

U

on a n y c o m p a c t

of a n a l y t i c

subset

has

in

subset

If

Let

I~)

sets

we m a y

V

find

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

U

obviously

have

any analytic

We c a n

of w h i c h

set

is a n e i g h b o u r h o o d

I~o

analytic

SnV ~AvnV v,

is an a n a l y t i c

each

open

S) Xv,

Proof.

are

a = O.

U n R p (A)

X~,

subset

analytic

such that

satisfies

of C h a p t e r

is an a n a l y t i c

hence

Av

suppose

that

a

of

an a n a l y t i c

and

components

analytic

if

We m a y

p

A = U --v,a A is the --a irreducible components, and

into

(A - A ' ) n U ,

p = d i m --a' A

S

is t r i v i a l . )

in fact,

S )A --a --v,a

[in the n o t a t i o n such

theorem

on the d i m e n s i o n

Vv9

We m a y U

induction

such

is a n e i g h b o u r h o o d

V

by

is i r r e d u c i b l e :

in an o p e n

V

proceed

> s o,

so t h a t

to

Aso.

hence A~nV

so t h a t we of

a

Since

--s,aA = --so'Aa = AaonV

if

100

Corollary Q,

2.

then

n As

Proof.

Corollary

is a n y f a m i l y of a n a l y t i c

is a n a l y t i c

from

3.

R n,

many

{As}

Apply Corollary

of e l e m e n t s

in

If

I to the

A

points

Let

isolated

points

f a m i l y of f i n i t e

is an a n a l y t i c

t h e n any c o m p a c t

Proof.

~. intersections

{As}.

If

isolated

in

sets in

K

of

s u b s e t of

~

contains

set

only

finitely

A.

be c o m p a c t of

set in an o p e n

A,

in

~

and let

and A

{cs}

the

family of

= A - {cs}.

Clearly

A~

by Corollary

I, t h e r e

S

is a n a l y t i c

in

~.

Let

B = ~

As;

S

exist

finitely many

s,

say

BnK = Clearly

c~K

if

~

s I .... ,Sp,

~ As. nK. i=I l

is not one of

N o t e t h a t the c o r r e s p o n d i n g analytic

sets

as above)

(which c a n o f c o u r s e be p r o v e d

(see C o r o l l a r y

b

(or

E, F

If

E~

K such that

or

F

III, A n-I

Er

K

such that

close

F)

to

for c o m p l e x in the III,

same w a y

Proposition

II).

R n,

d i m Ab = I

and

for

K a closed

~,

we write

if t h e r e e x i s t s EnU ~ FnU

or

an o p e n

subset E = K F

set

U) K

The

K-

EnU r FnU

respectively). Let

S, A

complement set

B

of

2 to T h e o r e m that

be a n a l y t i c

CK(S; D

A)

of

such that I,

B

S = K A~K(S;

subsets A

S

Further,

~, A r S.

is the

B ) K ( S - A).

exists.] A).

in

of

11

Further, there n in R , with A_o

O.

s u b s e t s of

(or

EnU = FnU

Proposition

set in

are two

of C h a p t e r

n e a r the o r i g i n

arbitrarily

b e an o p e n

Let ~.

sets

of d i m e n s i o n

some p o i n t s

of

to C h a p t e r

analytic

irreducible

s i, .... Sp.

proposition

is an e a s y c o n s e q u e n c e

e x i s t real

such t h a t

smallest

analytic

[By C o r o l l a r i e s

i and

it is e a s i l y v e r i f i e d

101

Definition

3.

any analytic we have

S sets

A ) S

Remark

i__ss K - i r r e d u c i b l e A,

or

B

of

B ) S

~

(on

t h a t it f o l l o w s

if

SnK ~ O

such that

and for

S = K AuB,

~). t h a t if

S r

AuB,

then

S cA

K

or

S (B.

and

S

Further,

is

a

I0. ,

itself.)

Let

is a c l o s e d

_Aa ) --aA. Then

A

then

S

set w i t h

is a l s o

K oK'

K'-irreducible.

A be an i r r e d u c i b l e g e r m of a n a l y t i c :a A b e the s m a l l e s t a n a l y t i c s u b s e t o f

and let

such that

with

K'

K-irreducible,

Proposition set at

if

(There is at least one

is

K-irreducible

s u c h set,

for a n y c l o s e d

viz. K

a~K.

Proof.

If

A =

is i r r e d u c i b l e

'

BuC, K either

then -B- a

B uC A , --a --a ) --a

) -A- a

or

-C- a

so that,

) -A- a ,

since

A --a

s a y the first.

A

Then

BnA

B n A ~A,

induces so t h a t

Corollary.

an i r r e d u c i b l e is

A

is an a n a l y t i c

then

for a n y o p e n

analytic

set

We m a y c l e a r l y

as above,

clearly

= --a A .

analytic

A

A cA,

is c l e a r l y

in

~,

neighbourhoods U

with

Definition A

A'

suppose

set in set

in

~a'

~,

Ur U

hence

acA a~U,

whose

o.f

S

of

and

A --a

there

is

g e r m at

a

Then

A

Let

i__nn ~

K-irreducible

statement:

and

"if

is

and c o n s e q u e n t l y

A

is an a n a l y t i c i

set

A

analytic

is i r r e d u c i b l e " ,

be an a n a l y t i c

A , & C K ( S ; A)

acQ,

a fundamental

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

then

S

if

irreducible.

a

A' = --a A , --a

U = ~.

~ a ( ~a'

and if t h e r e e x i s t s U

4.

that

so t h a t

N o t e t h a t the c o n v e r s e

in

containing

~a"

Proof.

~a

a

B ~A.

If

is i r r e d u c i b l e ,

a g e r m at

set in

A r S

is s y s t e m of sets

A'

is t r i v i a l .

Q. A K - c o m p o n e n t

such that

A

i_~s

102

Proposition ii. is any B

A

K-component of

is such a

of

Let

S,

B ~ A,

then

S.

B ~ A.

If

B

Moreover,

K-component if and only if it is a

Suppose

Since

Br A;

B cA r CK(S ; B)

B

K-component.

is a

also

K-component

B

be a

B r A,

is a

that is

and

latter union

B),

K-component of

S.

and we set

A

BtA.

Conversely,

Then

B# A,

CK(S; A).

we have

and

E = CK(S; A),

B

let

since if which

by our assumption Consequently,

B ~ AUCK(CK(S; A);B),

)CK(S;

we have

C_K(CK(S; A) ; B) = CK(S; A)

K-component of

Remark that if

since

CK(S ; A).

B #CK(CK(S; A);B)

K-irreducible,

A r CK(S ; B),

B ~CK(S; B), ; A),

CK(S ; A).

we have trivially is a

since B r

K-component of

K-component of

B

A ~ B.

this implies that

Further,

is not the can since

A / B,

contradicting the assumption that

B ~CK(CK(S ; A) ;B)

Hence

is a

then, since

A r S = K BUCK(S; B),

so that

B

K-component of

CK(S; A).

Proof.

B

be a

since

and since the

B ~ CK(S; B),

so that

are analytic subsets of

B

S

we have

CK(S; AuB) = CK(E; EnB). In particular, S,

if

A

and

B

are distinct

K-components of

then

CK(S, AoB) --CK(CK(S, A),B). Let now

{As}

be the family of

For each finite set

I = {a l,...,ap},

Proposition II, the

As,

s~I,

K-components of let

are the

S.

AI = s~I ~ A~. By

K-components of

CK(S; AI)Definition 5. CK(S; AI)

The

K-residue of

S

is the intersection of

over the family of all finite sets

I = {a i ..... ap}.

103

Since we have from C o r o l l a r y K-components

Then Aa

its

12.

S

the

B~

such

that

are

finitely finite

and

C r~

are

Proof.

meets

~,

J,

the

the only

finitely if

of

K

of its

B ~ is analytic of

S

if

~ ~ ~',

only

in

~

for for any

is e m p t y

~ CK(S ; AI) I={a i ..... ~p}

AanC

where

if

B~

In fact,

i to T h e o r e m

is empty.

K-components

B~ I B~,,

of

if

S

S = K U B~,

meets

K-residue

of

many

K-components

a~ I%

easily

(the union

K

K-residue

sets with

~kJ j

and

then

S =

Further,

subset

Corollary = @

that

K-irreducible

any c o m p a c t

set

i that

S = K U A a.

many

we d e d u c e

AI)'

K-residue).

Suppose

any c o m p a c t of

Bp

i to T h e o r e m

and

Proposition

S = K AIUCK(S;

and the

S.

i,

CK(S ; AI) nC = @

The other

statemen~

= ~, for

then,

some

by

I,

and

of the p r o p o s i t i o n

are trivial. Note

that

satisfied

if the

Proposition S

13.

is empty,

Further, where

_Aa

If

then

since

K

A

is

is compact,

finitely

K

many

K-components

UA

A

such

Proposition

subset

that

Let of

B / A,

of

the are

in

~.

K-residue

of

finite

A

in number.

(see P r o p o s i t i o n

--aS

at a point

• that

I0)

acK.

be a Then,

then

there

A such that --u,a i to it b y P r o p o s i t i o n IO,

13 follows

A ~.

from T h e o r e m

components

S =

a compact

form

component

it follows

irreducible

14.

S

family

A is an i r r e d u c i b l e c o m p o n e n t of S --a K - i r r e d u c i b l e b y P r o p o s i t i o n i0. Further,

is the set c o r r e s p o n d i n g

Proposition

of

is c e r t a i n l y

finite

then

is of the

Av,i

.

{B~}

is compact,

is an i r r e d u c i b l e if

on the

form a locally

K-component

In fact,

a,

{B~}

and the

each

Proof. at

the c o n d i t i o n

at once

K-irreducible if

B

are

if then

from P r o p o s i t i o n

set,

where

is an analytic

dim _Ba < dim --aA for any

K

subset a~K.

12.

is of

104

Proof.

By C h a p t e r

prove

that

A --u, a

of the g e r m of

analytic is a

B

III,

d o e s not c o n t a i n A --a

at

that

A

of

A.

= A,

any

7, we h a v e o n l y to irreducible

a~K.

set c o n t a i n i n g ~ u ,a'

K-component

implies

Proposition

If

then,

Since

A

A

component

is the

smallest

by Proposition is

13 '

K-irreducible,

and we m u s t h a v e

B = A,

this

a contra-

V

diction. This proposition set

K

by a closed

Propositions Bruhat-Cartan

Definition we

6.

~

Rn

if

follows,

in a s e n s e

set we

than,

There

A = {xs

a coherent Note converse

s t a t e the r e s u l t s Bruhat-H.

Proposition

are

finitely many = 0

is a c o h e r e n t / ~R},

relating

Whitney

long.

A

[6].

They depend

similar

to, b u t

9 above.

A subset

analytic

functions

for all

i}.

s h e a f of i d e a l s

i.e.

~c R ;

S n R n = A.

two

A = {x~Ifi(x)

There

such t h a t

if one of the f o l l o w i n g

is f u l f i l l e d .

that

~'

of

conditions

(ii)

in

n

set

A

if and o n l y

such that

set

[Ii])

of the

is c a l l e d n n ~'c C , Q'nR = ~

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

(H. C a r t a n

C-analytic

(i)

shall

open

A

s i n c e t h e y are r a t h e r

stronger

15.

the m a i n r e s u l t s

components.

Cn

sets o b t a i n e d b y F.

on p r o p e r t i e s

Proposition

S

the c o m p a c t

itself).

be a s u b s e t of an o p e n

exists,an

the p r o o f s

essentially

~

as a s u b s e t of

analytic

C-analytic

We o m i t

A

there

In w h a t

(e.g. b y

if w e r e p l a c e

12 and 13 r e p r e s e n t

Let

and a c o m p l e x

to

set

false

t h e o r y of i r r e d u c i b l e

look u p o n

C-analvti

becomes

is the

~ (~R

is

fi

such

set of z e r o s of

s h e a f o f ideals. that a c o h e r e n t

is n o t t r u e

analytic

in g e n e r a l .

set is

C-analytic.

The

in

105

Proposition

16.

If

A

is

C-analytic

is a c o m p l e x a n a l y t i c set A n C such that for any c o m p l e x h o o d of

~ ~

in in

Cn C

n

U

of

A

of any f a m i l y

in

~, then there

in a n e i g h b o u r h o o d analytic

for w h i c h

S )A,

set

S

there

of

~

in

in a n e i g h b o u r -

is a n e i g h b o u r h o o d

A

with

SnU ) A,~U.

{As}

of

Further,

C-analytic

the

sets

intersection

is again

C-analytic. It m a y happen,

however,

that

~ N Aa;

one

a l w a y s has

c NA a .

Definition

7.

~

C-analytic

not the u n i o n of two Note analytic

that a

Given

irredundant sets

irreducible

itself.

set m a y be r e d u c i b l e

C-analytic

locally

AS

with

18.

Let

if

dim B < dim A. p,

dim B < p

and e v e r y point of p,

dimension

p.

Note

that

i.e.

is a

A

dimension

dimension

A - B

that any a n a l y t i c

points

of

contains

f a m i l y of

as an

there

C

A 14)

is

C-

is

A.

C-irreducible

C-analytic

B cA,

Then

is a

C-analytic A - B

B ~ A. C-analytic set

real

Br A

manifold

analytic

set c o n t a i n i n g of

B,

A.

set,

set of

is a r e g u l a r

is an a n a l y t i c

the w h o l e

is a

C-irreducible

Further,

A = dim

for an a r b i t r a r y

happen

A

R

be a

if

A,

(see P r o p o s i t i o n

set w i t h

Futher, there

A dim

A

C-analytic

finite

set

A = U A S.

and we h a v e

Proposition a

a

if and o n l y

irreducible,

B

if it is

sets d i f f e r e n t f r o m

C-irreducible

17.

C-analytic

and

C-analytic

C-irreducible

set.

Proposition unique

set is

set

such that p o i n t of of

A,

it m a y

the set of s i n g u l a r

106

3.

Examples In t h i s

H.

Cartan

the

x,

[ii]

very

that

section

irregular

follows,

y,

and

we

we F.

shall

give

examples,

Bruhat

- H.

Cartan

bahaviour

consider

of

the

real

space

due

[5]

to

to

illustrate

analytic

sets.

R3

coordinates

with

In all

z. 3

Example

I.

Let

Ar

R

be

the

cone

is

irreducible

a neighbour~ood x = y = O, that

Example

2.

variable for

z,

-i

of

which

follows

< z

and

of d i m e n s i o n

(O, O,

z),z

2

~ O,

is of d i m e n s i o n

A

is n o t

Let

a(z)

= O

for

< +I.

z

Let

the

S

is

the

z set

However,

reduces

to

the

By P r o p o s i t i o n

7,

it

function

and

be

O.

although C~

~ -I

at A

I.

coherent, be

equation

= x3.

z ( x 2 + y2)

A --a

whose

~ +I, of

is of

in line it

C-analytic. the

real

= exp

points

( I / ( z 2 - I)) 3 in R

satisfying z ( x 2 + y2)

Then

S

is an

the

points

analytic

of

term

on

the

This

proves

S

is n e a r

that

S

[il])

vanishinq

on

such

identically Example by

3.

that

S

can

that

where

right

H. C a r t a n

One

set.

is

even

any

For

(y - i ) 2 =

z = ~

i.

0

which

any

real

> O,

(z - n ) x 2,

Cartan let and

to p r o v e

near

such

implies

that

It c a n

be

this

points,

near the

x = y = O.

shown

function

(see 3 on R

zero.

compact

function

(see H.

But

analytic

identically

construct

n

is e n o u g h

is a n a l y t i c .

analytic

zero

It

= x 3 a(z) .

analytic

vanishing

sets

on

S

S

in

R

3

is

[II]). Sn let

be

the

set

in

R

P

be

the

plane

3

defined x = O,

107

and

let

A =

U

SnuP-

Then

A

is an a n a l y t i c

set

(of

- P, n meets

we

n)o

dimension have

z

only

for

A

2

at e v e r y

) n

so t h a t

finitely

point).

In fact,

any compact

many

n.

set

However,

on

S 3

in

R

the

singular

points

contain

ix = O,

the set x = O, y , z ~ n but not 1 1 n ' y - n' z < n~.j S i n c e the s m a l l e s t a n a l y t i c

taining

all

I

of

the

sinqular

analytic

points

of

4.

as

y - n , z ~n

x = O, A

set c o n t a i n i n g

same dimension Example

sets

is n o t

analytic

the

singular

the

irreducible

let

Let

Note

D r A

that

A

be

motion

outside Let

let

be

Dk,

the

I =

denote

= [ n i.

then

the

TI(A ) = A I

However, S = BuC,

TI TI

does

(and in fact,

any

of

cone

A

in

and

S

has

C ~ S,

in

has

the

R 3,

A'

= A - D.

as f o l l o w s . is so c h o s e n

no

chosen;

e.g.

if t h e y

T I ( D k) ~C,

TI(D) r

B ~C,

in

I

set of d i s t i n c t We d e f i n e ,

I,

If

locally

finite)

~A I

I.

Then

component.

and

S

In fact,

hence

Dk

(if t h e on

A'),

for

infinitely

all

(since

reducible.

for

lines

sphere

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

hence

let

2

~ Tj(A') ) p (J)

O, n~z Since

u~'

principal

R'

R'

u~(~') 2

Then the principal ideal ~'

be holomorphic

function~ in fi

S1 .

(The

small enough.) Now, the germs ~R'

there exist

qi

f, ~i' ~i

does

not

=

qi(b) ~ O 5(Sl),

can be found if

U

is

so that there

V on

and, for

x

i. near

f a]. ga

=

V rU

of

such that S1

Then, b,

f

a

and holomorphic

induces the germ

and such that

a

on

be a point at which for each

such that

[u

~i fi = ~i f b~SlnV

of holomorphic germs

~i~'

~ifa ,

on

vanish

(2.1) Let

Sl,

u = (fi) a"

there is a neighbourhood

functions

Let

U, vanishing on

(fi) a e ~ r

~i' q i ' % '

qi(fi) a

fa'

R').

such that ai

Hence,

u R' = ~,n,

we have

vanishing on

Hence,

into

is the only prime ideal in

on

~

R'

[which exists because

5(S I)

exist

~'

is a Dedekind ring, and

which generate the ideal sheaf U

and

~' = u R', so that R' is a fa ideal domain. Let u - g , fa, ga~Oa ' ga$~"

fl .... 'fk

,

R'.

can be factored uniquely,

u~',

(since 2

R,

is

the only non-zero prime ideal in

~ '. Hence, by definition,

~' # (O), R').

R'

This is a local ring with maximal ideal

noetherian and integrally closed, and hence~so is Since

Let

V.

S1

since fi

is regular and fl ..... fk

generate

is a multiple of

f,

ii7

we

see

hood

that

the

b.

We

of

a regular

Thus,

3.

Let

b~X,

let

X

on

be

have

since

~.

is n o w h e r e at

x~X,

hs

any

regular

Proof.

Let

generate

U

the on

the

sheaf

X.

vanishing if

z~A

Hence

hi, . . . ,h m point

Theorem

3.

that

Proof. Theorem the

Let 2,

of

Let X

X

A x, of

set

A

of

~r

C

let

point

of

i ~ j

~ n

vanish

be

of

g

is

g = h(f)x , where

is a r e g u l a r

f,

point

of

h i ..... h m

generate

functions

IV,

Remark

after

Theorem

A,

if

= f,

we

h

o A

=

n

-

in 6,

have

p

+

i.

--Z

and

1 ,<j ~ n on

Xz

m

it

follows

that

z

is

a

'

a complex

space.

a

normal

at

a;

then

U

of

a

dim S b

X.

functions

of h o l o m o r p h i c

at

X,

,

f

X.

be

n

zeros

if

of h o l o m o r p h i c

By C h a p t e r

is n o r m a l X

set

>~n - p = d i m Xz,

X

b

to p r o v e

and

that

then

A

of g e r m s

{ m,

only

set

that

is a c o n t r a -

point,

and

hypothesis,

a neighbourhood

singular

the

3 below)

an o p e n

= n - dim ( i

regular

such

in

~(X)

J 1 ~ i ~m, since

that

(---~.11

rank

in

a neighbour-

holomorphic f u n c t i o n s on a n in C , v a n i s h i n g on X, which

By our

a regular

0h. (--~) 3 o

rank

set

in

be

sheaf

A.

have

at e a c h

on

point

b

~(A)

on

is

of

ideal

vanishing ideal

g = 0

Lemma this

2, w e

p

dense

hl,...,h m

neighbourhood

from

analytic

Suppose

holomorphic then

Theorem

an

~(S I)

bs

dimension

on

X

generates

(as f o l l o w s

X;

X

holomorphic f

of

f

to p r o v e

Lemma

of

claim

point

diction.

be

function

we

is o p e n

The

in

set

- 2

for

is,

that,

have

~ dim _~

aeX

X.

there

such

of

b~U.

by if

S

is

is

118

By T h e o r e m

i, the sheaf

is a s u b s h e a f ~a

= ~a .

~U %

Since

equals = %

at

of

~U'

~U

and since

b

near

subsheaf a; h e n c e

4.

an a n a l y t i c

subset of

for a n y

Let

aeX.

be e x t e n d e d Proof.

U;

further

is n o r m a l

~U

at

is open, b,

%

a,

the c o h e r e n t

for such

This

definition

X

be a n o r m a l X

sheaf

we have X

is n o r m a l

is an i m m e d i a t e

12; T h e o r e m

3. The n o r m a l i z a t i o n

theorem

Proof. point

space.

Because of

X

over

%' f. l hi _ gi'

and

V

clearly

set in

let

cn.

V

X - Y

points

= [x,hl(x)

an a n a l y t i c

closure

Y

Y'

V • Cm.

To p r o v e

U

Oka)

can

III,

and the

that

seen

are h o l o m o r p h i c

gi

is n o w h e r e

of

V;

the

X

be

(Y,~). that a n y

U

X

admits

is an a n a l y t i c

(Lemma 1) that

h i,...,hm ~a

Let

Let

to p r o v e

such that

suppose

We have

enough.

3 above

a normalization

be a n e i g h b o u r h o o d

fi" gi

of e a c h

of

on

of C h a p t e r

of

2, it s u f f i c e s

let

is small

h(x)

theorem

has

H e n c e we m a y

where

set of r e g u l a r if

X

~a-m~

set of zeros

Y

spaces.

a neighbourhood

a normalization. set in an open

Then

of L e m m a

has

is a finite

function

consequence

(The n o r m a l i z a t i o n

any complex

and

dim --a Y ~ dim --a X - 2

2 and P r o p o s i t i o n

complex

4.

space

X.

of n o r m a l

Theorem

complex

such that

Then any h o l o m o r p h i c

to

Proposition

map

on

b.

Proposition

V'

X

the set of p o i n t s w h e r e

the c o h e r e n t

for

is c o h e r e n t

generate of

a

on

V,

dense.

Let

~a

~a

in w h i c h and the V'

be the

h0 are h o l o m o r p h i c on i m h : V' ~ V x C be the

. . h (x)). The image h(V') is #'#m set Y' ( V' • Cm . We c l a i m that the in

this,

V • Cm we m a y

is an a n a l y t i c suppose

that

V

subset

of

is i r r e d u c i b l e .

119

Let

x Cm

Yi r V'

~x,

be

the

z I .... ,zi_ I, h i ( x ) ,

z I,. ~.,9 i .... ,Zm

are

analytic

z i + I .... ,Zm)

arbitrary,

Y! in V x Cm . C l e a r l y l any a ~ Y i. Further, the

set

dim

Hence,

by Chapter

Y' l

in

V

Z. ia

x Cm

V • Cm

dim

=

IV,

in

V x C

closure

of

Y'

Y

Proposition

a;

the

hence

is a c o m p a c t is a p r o p e r analytic also

V

hi~ if

map.

V

set of

V,

same

and

Z. ) Y.. l l

4,

the closure Yi m Z = i~=I Y'~" T h e n

of

• C m)

the

= Y'.

is a n a l y t i c

Z

Hence

(by C h a p t e r

of

Cm

{x}

enough, Hence,

if

equation

and that

with

where

K

the projection ~ : Y ~ V -I ~ (x) is a c o m p a c t

x~V,

• C m,

Y r V x K,

so is f i n i t e . every

hi

coefficients

(One m a y

satisfies

a

holomorphic

on

e n o u g h . ) F u r t h e r , if S is t h e s i n g u l a r -I A = ~ (S), then V' = V - S and

Since

h

is h o l o m o r p h i c

isomorphism.

Further,

on

V',

to c o m p l e t e

the p r o o f

only

to p r o v e

that

is s u f f i c i e n t l y

of

Since

every

fibre

hoods

saturated

if

a,

then

of

~

with

has

respect

W ~

-I

(W)

is a n o r m a l

~,

: Y' ~ V' Y - A

of T h e o r e m

a fundamental to

~

by definition,

Y. Thus,

this

is n o r m a l

at a n y p o i n t

Yo

with

is

4, we

small complex

system follows

space.

of n e i g h b o u r at o n e

3 and Y

IV,

in a n e i g h b o u r h o o d

in

Theorem

for

property

a~Z)

are b o u n d e d

is s m a l l

of

and

neighbourhood

(3.1)

they

of

is s m a l l

is an a n a l y t i c

have

a,

the c l o s u r e - i a defined by

for

Zn(V'

this by observing

Y - A = Y'.

dense

- I

x C TM

Further,

polynomial

if

V

V

subset

subset

prove

monic

in

the

Let

and

and

4') .

Since of

has

a

xeV',

= d i m m V • Cm

Proposition

n

Y.l

Z i cV • C

is a n a l y t i c .

is a n a l y t i c

where

and

d i m ~Y.

Zi = {(x'z) I gi (x) z'l = f'l(x) } (that

set

~ ( y o ) = a.

from

120

Proof of

(3.1).

Let

yo(~

be anal y t i c

h(x)

exists

X~VvnV' We cla i m that

z

the function

(~)

~ z

g = i

an e l e m e n t

Ik z l (v)

=

'"

on of

.. ' z(V)) m

'

w h i c h proves it is clear

that

if

V~,

= 0

on

he nc e a' we have

(a, z (v))

of

a

z (~) ~ z (u)

projects

into

this m e a n s

that

(y) = x]

near

by P r o p o s i t i o n

This proves

(3.1)

Y

near

near

a, = 0

on

Yo'

and w i t h

Let

seen that

Let ~X

X

uihi,

so that it T h e o r e m

be any c o m p l e x

sheaf of

(Y, ~

on the set of -I ~o~ is

a;

if we set

VN,

N ~ v,

ui~a.

above or directly).

~

d ef in es

But c l e a r l y

is normal

at

Yo"

4.

and

fu nc ti on s

~X on

the X.

Then

OX-mOdules.

[either

Chapter

Y

space,

be a n o r m a l i z a t i o n

= ~* (~Y)

~x-coherent.

D

and is a n e i g h b o u r h o o d

sheaf of qerms of w e a k l y h o l o m o r p h i c

Proof.

2,

[ ui(~(y) ). zi, [if y = (x,z I ..... Zm), i=l (u) Yo = (a, z ) on Y. Hence ~ is

on

is a c o h e r e n t

If

small n e i g h b o u r h o o d

V'nV v

~ =

u i ~ a.

g

~ =

holomorphic

4'

Hence m

Then

v~,

V . Let n o w e be h o l o m o r p h i c u points of D n Y and bounded; then

~a"

consider

v / ~.

on

an ele m e n t of

is

V v,

Further,

Vv

v

(x,z (v))~Y.

In fact,

0 = [ ui(a) z(U) i '

h o l o m o r p h i c and b o u n d e d on -I = ~o~ on V ' n V v near

OX

Hence

g = [ uihi,

that any s u f f i c i e n t l y

of

Theorem

N / v.

V

Proposition 2,

v.

for each

(v)

I = [ u i(a) z i(N) '

regular

U --v,a V be the components, and

of

z (u) = lim

z(~)

(a)) ~ a

V into i r r e d u c i b l e --a By sets r e p r e s e n t i n g V * --D,a

decomposition

defines

-I

of

X.

It is e a s i l y

from the proof of

IV, T h e o r e m

7 implies

(3.1) that

given ~X

i2i

Theorem of

5.

x~X

Let

at which

subset of Proof. %/~x

X

~ O,

Chapter

X

is not normal

the set

i.e.

IV,

Proposition

5. This

at any

dense

than

dues

for which

Since

~/O

is a

set by

analytic

Let

seen.

u

Since

C l e a r l y we have an inclusion ~

~

By Chapter

on

U,

U

on ~

5

U).

of any

S

is a

be the sheaf

which vanish

We claim

III,

is a universal

= O};

is coherent,

~ c~. (on

U

U

which

and let

functions

3 = Hom ~ (~,~).

5.

S = {x~Ulu(x)

set in

of

for the construction

is a n e i g h b o u r h o o d

function

x~U.

injection

x(X

to Grauert-Remmert,

sufficient

5, there

of germs of holomorphic

a natural

analytic

is an analytic

proof of Theorem

to Theorem

denominator

N

as we have

and a holomorphic

and let

(closed)

N

6.

is more

The Grauert-Remmert

nowhere

is a

is the set of

a proof,

of the normalization,

aaX

N

sheaf of ~ - m o d u l e s ,

Corollary

Then the set

N = {x~X I (~/~)x ~ O}.

We end by giving Theorem

space.

X. Clearly,

coherent

be a g o m p l e x

on

so is ~

that there

In fact,

~S, . is

let

%

fa~x be a non-zero divisor in ~s Hom~) (Jx' J• and let x is ~ - l i n e a r , we have a(g) = wg and w = ~ Since c~ f x lies in the complete ring of quotients for g~J . Clearly w x (containing since ] x is a finite ~ -xm o d u l e x N is integral over ~ x ' so that, b y non-zero divisors) , w Proposition

i,

We assert, of

X,

w s x. that

Thus, if

N

we have

satz

If there

'

~. points

we have

xeNnU;

Let

~c~=

is the set of non-normal

N~U = { ~ U I Ox

Proof

inclusions

t en then is

~x ~ % "

k >~I

%

~ F~}-

= Now,

such that

a fort or by the Hilbert ~k ~ x

r Ux~x c % x

%

= ;x

Nullstellen(since

u

122

is a universal denominator at that

~-I

~x ~ %

Then, we~x as yeU - S

and and

bounded near that w ~ ~x"

x.

of

~x

%

~ %"

since

x).

~x ~ % "

Further, we may suppose Let

w~ x "r now if ~e~x, y ~ x, sLnce ~(x) = O since Hence , if @~x'

Thus the map ~ f ~ w f

into [tself, and, since

W~x,

Y

-i % ,

w ~ x-

then (w~)(y) ~ O and w is w~e~x, it follows

is an ~ x -linear map a~%,

so that

Thus

= {x u r ([/e)x SLnce

w~

o}.

is coherent, this set is analytic by Chapter IV,

Proposition 6. Theorem 5 is proved.

123

CHAPTER

VII.

- HOLOMORPHIC

In t h i s R.

Remmert

analytic of

and

The

K.

set, a n d

Remmert

space

chapter,

[28,

under

last

shall

Stein then

30]

[32]

use

prove

on

first

the

asserts

holomorphic

sections

OF COMPLEX

contain

a theorem

a proof

that map

some

SPACES

singularities

it to g i v e

which

a proper

two

we

MAPPINGS

the

of

of

of

an

the

image

theorem

of

a complex

in an a n a l y t i c

applications

of

set. these

theorems.

I.

The

theorem

All

complex

countable

Theorem

~ p

the

is

dim

Z

and

for

aEA.

[dim Y

any A

the set

Y.

A

analytic b~A m

of

(Am

of dimension

to b e

Suppose,

all

analytic

Z

be

is

sufficient

is

analytic

Y,IX'

the

set

with

~

Let of

if

m).

of

A =

~ - Y

is t h e u n i o n It

means an

on

analytic X,

X'

a~Y,

of

of

an

the

suffices

a,

Am , that

where

irreducible

Y,

cl~ure

that of

= m

the

of

that

analytic Y C

is

is

n

of

an for

components

that

theorem

suppose

an

Am

dim b Am

that

induction

we may

subspace

to p r o v e

the

(by o u r

by

X.

with

that

can,

the

points

suppose

affine

~ n~m~p such

we

in

suppose

Z c D

to p r o v e

Y = @.]

that

singular

that

while

set

we may

subsets

in

Suppose

that

moreover,

of

an a n a l y t i c

Y ~ p - I,

is n o n - s i n g u l a r ,

Again,

Y

X - Y.

dim

is

local

a neighbourhood

i < p. subset

i__nn X

~c C n.

X'

Since

A

and

~ -I

is

It

intersection

dimension

assumed

of

that

theorem

for

is n o n - s i n g u l a r .

the

of

Let

in

automorphism

subset

such

proved

hypothesis). Y

an a n a l y t i c

= ~ - Z.

of

are

a comple x space

p ~O

an o p e n

< dim

be

inteqer

Since

X'

A,~X'

A

closure

is a l r e a d y

considered

~.

X,

is an

Proof. X

spaces

X

of

dim a A Then

at

Remmert-Stein

i.

subset there

of

every of

A

closure

A

m

124

of

A

is a n a l y t i c .

m

(l.i) with

If

~r C

an a f f i n e

A r~ - Y beA, a

subspace

for a n y

such that

AnU

To p r o v e

II,

is o p e n ,

and

of

is an a n a l y t i c

then,

that

n

Thus we have

C n,

there

m

{x~

(ii)

For

a

point

Proof.

Let

constant

on a n y

=

... = i

(countable)

Let

{x~}

linear

of the 11

be

form,

positive

be

set

for a n y U

as

irreducible

a countable independent

follows.

We may

clearly dim

repeat

dim

~< m a x

dense of

suppose forms

this

C

n A

of

X

and

is n o t

component = o}

~ max

= 12(x)

countable

many

=

values

for w h i c h

12

a

constant

on a n y

of a n y of t h e

and

sets

for w h i c h

(0, d i m Y - 2) .

= ll(Xu) , 1 2 ( x ) = 12(xv) } ,< m -

change

e(z)

= Im(Xv) }"

is n o t

and

= O}

m

~ m - 2.

We have

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

IZm+ll

only

to

times. we may

suppose that n in C , we have

z are the coordinate functions 'n ..... Im = Zm. Now, B = AuY is c l o s e d

function

is an

(0, d i m Y - 1)

z I .... z_l

xu

which

of

subset

{x~AIll(x)

construction

By a l i n e a r

is c o u n t a b l e .

{xu} cA,

II, w h i c h

{xs

{ x ~ A l l l(x)

set

f o r m on

irreducible

= l l ( x u) },

= O}n(AuY)

component

d i m Y n l ~ i(O) n12 I(0) Then,

(x)

m

of

{ x ~ A l l i ( x ) = li(xu),...,im(X)

a linear

dimensional

{x~AIli(x)

the

> 1

set.

dense

dim Ynlll(O)

=

of

and

is a n e i g h b o u r h o o d

we proceed

I ll(x)

isolated

11

l,

dim b A = m

is an a n a l y t i c

(I.i),

intersection

of d i m e n s i o n

set w i t h

aeY,

the

following.

There exist linearly independent linear n C w i t h the f o l l o w i n g p r o p e r t i e s : on

(i)

if

Y r

the

a = O.

. . . ,i

and

to p r o v e

2

on the

+

... +

set

_~zI

IZnl .

2 o

can .

in

Q.

take

Since only

= Zm = O } n B ,

2

i25

there

is

6

> 0

(arbitrarily

small)

such

that

B n K = @, where

K

is the

set

Since

K

is c o m p a c t

{z I = and

... = Zm = O,

B

closed,

I Z m + l l 2 + . . . + I Z n ~2 = 6}.

there

is

2

{lZll

,< ~ . . . .

Let

U

be

the

and

U"

= {z"r

BU = B n U

set

= ~(Y)

An(U

Corollary of

is the

- Z)

= closure

b

b~A. such

means,

in

Then,

x A,

on and

o f the

above

theorem

complex exist

A the

intersection

of

U'

- Z)

is an

with

of

theorem

Let

Hence,

by

the c l o s u r e

A,

U.

there

are

Hence

so t h a t

Toprove

points -1 x x(x).

of

3 in C h a p t e r

is c o m p l e t e l y

immediately

in

x, h e n c e

~(y) ~Y'

a

is p r o p e r ;

III

in

point

3 to T h e o r e m

U),

X

U.

is a n a l y t i c

isolated

< ~},

is p r o p e r .

Z ) Y.

is d e n s e

by construction,

of

with

~ i < m.

in

that

if

subsets B U,

is a n a l y t i c

- Z)

Then,

- Y'

such

< ~ .... , I Z m ]

to

since

> O

= 6}nB = @.

restricted

10 in C h a p t e r

in a n e i g h b o u r h o o d

We g i v e

The

x

< 6}.

of d i m e s i o n

An(U

by Corollary

is o p e n near

An(U

that

{z'~cmllzl I

for o t h e r

= B n ( U - Z)

U

that

of

=

2

+...+IZnl

x : B n ( U - Z) ~ U'

- Z)

in

We c l a i m

let

Cm

3 to P r o p o s i t i o n

An(U

U'

notation

~ : U ~ U',

l i n e a r s u b s p a c e of -1 Z = ~ (Y'). Then moreover,

x U",

similar

the p r o j e c t i o n Y'

U'

,< c, IZm+ll

I IZm+ll 2 + . . . + i Z n l 2

(with

Further

,IZm~

e

there An(U

A =

this, x~A

near

But t h i s

III,

that

xl A

are

points

y

- Z)

is d e n s e

proved.

an a p p l i c a t i o n ,

due

to

H. C a r t a n ,

theorem.

of C h o w .

projective

finitely

many

Any

space

analytic

~N

such

A

is an a l g e b r a i c

homoqeneous

Pl ..... P k ~ C [ Z o ..... ZN]

subset

polynomials

that

of t h e set,

i.e.

there

126

A =

Proof. whose in

p1(z

Let

X

images

C N+I -

we h a v e Hence,

,

belong

in

X

of

C

N+I

- {0}

is an a n a l y t i c

for a n y

z~X

X

1, the c l o s u r e

f '

D

A;

= o}.

and

A ~ O,

at a n y p o i n t

X = Y

of

set

X

is

in

C

A~C, > O. N+I

set.

fl ....

cyl inder

to

the d i m e n s i o n

by Theorem

Let

)

and since,

Az~X,

is an a n a l y t i c

=

be the set of p o i n t s

in ~ N 0

)

be h o l o m o r p h i c

functions

in a p o l y -

m

about

0

s u c h that

YnD = {z~D I fl(z)

=

=

f

m

(z)

=

o}.

Let CO

f'l(Z) =

[

Pvi(z)

v=o

be the e x p a n s i o n nomials

in

degree

v.

D;

We a s s e r t

of P

vi

that

Z = {zEcN+I IPvi(z) Clearly

that

a

is

here

= 0

Let

for all

kz o ~YnD ;

hence

CO

polynomial

v ) O,

polyof

Y

i = I ..... m}. and

Z

are

b y a c o m p l e x number,

that

Zo~YnD.

of homogeneous

a homogeneous

and since b o t h

To p r o v e

ZnD ) YnD.

series

where

by multiplication

Z r Y.

Z ) Y,

Then,

for

it s u f f i c e s A~C,

lAi

left

this

implies

to p r o v e ~ 1,

that

we h a v e

CO

[

Pvi(AZo)

=

V=O

Clearly

in

Y = Z,

ZnD ( YnD,

invariant

f.

this

Since

[

AVp

(zo)

= 0

for

Pvi(Zo)

= 0

for e a c h

D =0

implies

that

C[z o .... ,ZN]

many homogeneous

vi

is N o e t h e r i a n ,

polynomials

i = 1 .... m Ikl " '

,< 1.

v, i = l t . . . # m .

there e x i s t

Pl ..... P k ~ C [ Z o ' ' ' ' ' Z N ]

finitely such that

127

C n+i I P i (z) = O for all

Y = This

u, i} = {zecN+llpl(z)

= . . . = pk(z)

= O}.

is just the theorem. nk

Corollary.

Any analytic

algebraic.

This

follows

and the r e m a r k

that ~ n

N =

- i

(n+l) (m+l)

subset of

at o n c e xpm

~Ivnl• ... •

from the t h e o r e m

becomes

by means

is of C h o w

a submanifold

of

~N,

of the Segre m a p p i n g

(xO ..... x n) x (Yo .... 'Ym ) ~ (XoY O ..... X o Y m , X l Y O ..... x l Y m ..... X n Y O ..... X n Y m) For a q u i t e d i f f e r e n t deeper

s t u d y of a n a l y t i c

see Serre

[24,

holomorphic Let

X

= k

theorem

be

map

a holomorphic

jacobian matrix

X

If

of of

X

= dimxX

f 9 X - + Cn

points

of

into a c o m p l e x

1.

is R e m m e r t ' s

of a n a l y t i c

sets u n d e r

X, f

ex(f) =

max xEX'

complex

x~X.

Let

space Y.

space,

proper

and

map, p

is a c o m p l e x m a n i f o l d .

be a

We set

- d i m x f-lf(x).

let

X'

complex

be a d e n s e

the m a x i m u m

at any point

Qx(f).

i.e.

f : X ~ Y

is a pure d i m e n s i o n a l

By the s e m i c o n t i n u i t y max x~X

section

a pure d i m e n s i o n a l

Proposition

have

varieties,

mappings.

Qx(f)

Proof.

of this

25] on the images

is i n d e p e n d e n t

holomorphic

regular

of a l g e b r a i c

and a m u c h

mappinqs

The principal

dim x x

properties

theorem,

[35].

2. H o l o m o r p h i c

theorem

p r o o f of this

of

X'

subset

Then

p = max xs IV, we

we m a y s u p p o s e

The c o r o l l a r y

of

of the r a n k of the

t h e o r e m of C h a p t e r Hence

space

follows

then

that from

Qx(f) .

128

the r a n k t h e o r e m

(Chapter

I) .

We call the i n t e g e r If

m a x Qx(f) = Q(f) the rank of f. x~X is not pure d i m e n s i o n a l , Q(f) = m a x Q ( f l X ), where

X

X =~Jx~

is the d e c o m p o s i t i o n

ponents.

(When

X

of

X

into i r r e d u c i b l e

is p u r e d i m e n s i o n a l ,

com-

the two d e f i n i t i o n s

agree.)

Proposition

2.

the c o m p l e x

space ont____oan a n a l y t i c

in

Cn

Proof.

Let

Then

f : X ~ Y

Q(f)

such that the j a c o b i a n

Now,

let

Hence, Yo

isomorphism open

set

Qx(f)

m

matrix

We Cm,

Thus,

m a p of g =

X

is c o n s t a n t .

is a s u b s p a c e

gl

factor }, of

X

U

of

and Yo

for any

and

irreducible

Then,

c # cv

if

c~

for any

of d i m e n s i o n

~ Q(f). h

an a n a l y t i c Y

onto

that

if

< d i m X,

an

g : X ~ W m

We C ,

then

{c }, v = 1,... components

{projection u,

Xo~X

xEf-l(u),

set

let

= dim Y.

point

on

to prove

set

has r a n k

dim Y

o n t o an o p e n

on those

C n)

xo

Y,

Then,

(gl,...,gm) ,

on w h i c h

first

into

at

of

we h a v e

of

on the

point

m = dim Y.

be the v a l u e s gl

f

m a p of

in an open

is a r e g u l a r

of

of a n e i g h b o u r h o o d

Let

Y

b y the r a n k theorem,

= Wx(h o f).

~ ~(g).

I, there

be a r e g u l a r

is a h o l o m o r p h i c

set

(as a m a p p i n g

B9 P r o p o s i t i o n

p = Q(f).

be a h o l o m o r p h i c

then

of

X

of

W -I Xl = gl (c)

and

,. ) : X 1 ~ Cm - I maps X 1 o n t o an o p e n s u b s e t g' = (g~ "''gm of Cm- . It f o l l o w s (by induction) that t h e r e is a r e g u l a r point

xl~X I

m - l ~ dim(Xl) where

gl

of

XI

for w h i c h

- d i m x l g , - I g , (x)~ d i m x l X - i is the r e s t r i c t i o n

component

Z

m

and the p r o p o s i t i o n

~ Q(g),

of

X

with

of

dimxl

g

- d i m x l g - l g ( x i)

~ Qxi(gi~l,

to an i r r e d u c i b l e

Z = dimxl X(xI~Z);

is proved.

hence

129

Theorem

2.

Let

X

_a p r o p e r

holomorphic

Proof.

If

is a g a i n

proper.

given in

point

C N)

finite

of

family

proper,

of

S

suppose

be

induction,

the

A = f-lf(S)

is

an

is a n a l y t i c ,

if

A

= X,

Then

dim

that dim

A ~ X. f(A)

such

= dim

that

Since as

the

X

our

the

fibre

of

of

flA

f

: A ~

of

f

by

an

Proposition

in

a set

in

CN

(U)

has

(X',

Y')

family

{Xa}

of

It

Hence,

the

suffices

subset

Since

the

is a g a i n

and set

let in

n

p = Q(f). ~.

Since

Hence

f(A)

is p r o v e d .

be

[such

Cn

Y.

induction,

analytic.

X.

a

C

in

pairs

By P r o p o s i t i o n

: X ~

~ -1

finite.

X,

theorem

Let

Y

irreducible.

in

< n.

and

is

set

of

set

finite.

analytic

set

= Qa(flA)

Q(flA)

the

: X ~

an o p e n

Y,

all

a closed

is an

A

for

is

in

suppose,

locally

set

analytic

of

fibre

that

T = Q(flA).

Q(fIA)

definition

to

in

We

f(X~)

map

set

an o p e n

locally

each

= T

X.

is

singular

f(S)

is

f

: f-1 (U) ~ U

f

a neighbourhood

compact

X

and

analytic

then

,

Further,

is

that

is an

Y

proved

of

spaces

analytic Y

< n.

a proper

we m a y

Let

X'

{ f ( X a) }

to p r o v e

restriction

that

already

dim

sets

therefore

in

n = dim

components

of

set

a closed

Let is

complex f(X)

is r e l a t i v e l y

with

irreducible

By

U

be

(imbedding

suppose

theorem

spaces

Then

as

dimension.

the

map.

Hence

yo~Y

if

Y

is an o p e n

we may

Moreover,

that

U

and

= f(S)

Suppose

then

2, we h a v e

a regular

point

a

by

exists

of

A

the

i]. through

a

is the

a neighbourhood

of

same

A,

we

have Qa(fIA)

= d i m a A - d i m a f-lf(a)

Moreover, because an

since

of T h e o r e m

analytic

Since

f

X - A

suffices

subset is

to p r o v e

is p r o p e r , i, of

it

we

suffices

~ - T

irreducible, that

< d i m a X - d i m a f-lf(a)

f(X-A)

have

f(X)

to p r o v e

is

of

- A) .

that

f(X

p

at e v e r y

of d i m e n s i o n because

= f(X

=Qa(f) ~Q(f)=p.

Proposition

analytic

in

Thus,

- A)

2,

~ - T.

is

point. it

130

Let of

the

B r X - A

jacobian

analytic f

~-

- T.

If

Hence f(X

dimension and

2.

Theorem

3.

Proof.

the

is c o m p a c t

for

sets

is

then

closed,

is

x o.

Let

whose

X - Ku Hence,

{x v}

and

result

I,

of

an

in

is p r o v e d . we

I)

have

= Q(flX-A-C)

to p r o v e

that

Q - T - f(C)

rank

rank

is

an

analytic

case

f

Suppose

is

X,

of

Yo"

is

that

is

of

be

f

Kv

: X-~

Y

Y.

then

that

is

it

f - l f ( x o)

false,

and of

a fundamental is n o t dense

is n o t with

for w h i c h Theorem

f

irreducible.

is n o w h e r e it

proves

a sequence

{Vv}

is

is an

constant,

this

Since

xvcX-

Hence

in

to p r o v e

(since

set

set

X

and

~

This

is n o t

sufficient

this

spaces,

when

if

p,

theorem.

complex

is a c l o s e d

a contradiction.

is

set

above,

be

each

there

Since

as

constant

f - l f ( x o)

for

< n.

analytic

{Kv} , K v c ~ v + I

union

B

rank

constant, in

X

compact,

by

f ( x v) ~V v -

f({xu})

3 is p r o v e d

{yo}.

is n o t when

X

irreducible. The

and

it

Clearly

Y

that

irreducible,

assumption).

is

assert

each

the

the

: X - A - C -~ ~ - T - f(C)

the

of n e i g h b o u r h o o d s

and meets

But

f(X)

fact,

is an

subset

the

and

at w h i c h

(by P r o p o s i t i o n

has

first

= f(Xo)"

X

< p.

of T h e o r e m

f

of

X

we

In

and

= Q(f)

j acobian

Then

is p r o p e r .

compact

is

(X - A),

since

Let

case,

Yo

of p o i n t s

of d i m e n s i o n

because

Consider

In t h i s

system

- A)

But

a close d map.

let

< dim

consequence

Theorem

f

is an a n a l y t i c

p.

immediate

of

X - A

suffices,

- A - C)

proper

C

< Q(flX

it

set

is p r o p e r , f(B) -i C = f f(B) = X - A, dim

f(C)

of

the

T

Otherwise, dim

matrix

subset

: X ~

be

the

general following

case

of

lemma.

Theorem

3 follows

at o n c e

from

this

= p.

131

Lemma

I.

and

Let

X,

f : X ~ Y

yo~Y,

there

subset

K

Y

be

locally

a continuous

compact

closed

is a n e i g h b o u r h o o d

of

X

such

map.

U

of

Hausdorff

spaces

Then,

any

Yo

for

and

a compact

that

f(K) nU = f(X) nU.

Proof.

If t h i s

is false,

of n e i g h b o u r h o o d s subsets such

of

X

that

in

X

whose

image

Let

any complex

space

point such

x has

a~X that

Proof.

components (for all

x

be

and

point

at

assert

X , and --a near a on

f

of

Theorem

set

3.

complex

a holomorphic

system

- Kv,

a contradiction.

it,

independent

xv~X

is a c l o s e d

a pure dimensional

is a n a l y t i c

is a r e g u l a r

closed,

system

of c o m p a c t

is a p o i n t

{xv}

f : X ~ Y is

We

there

and with

a fundamental

a~X.

of

X

a sequence

then

is n o t

lemma,

is a f u n d a m e n t a l

{Kv}

But

Y

f-lf(x)

f(U) Let

and

{yo}.

in

3.

k = dim

{Uv}

K u c K u + i ~ X,

the

Proposition

x

with

proves

that

Yoo

f ( x v ) ~ U u-

This

Y

of

and

of

x~X.

space,

map

such

Then

any

of n e i g h b o u r h o o d s

U

f(a) .

that

if

X

are

the

irreducible

f - l f (x) = k t h e na dim = fl~ , --v, v v,a x v v Xu). In fact, dim f~Ifv(x) = k if X

on

X

.

Hence,

by

Proposition

i,

V

f~Ifv(x)v

dim

) k

for

all

x ~ X v.

In the

other

hand,

since

X

f-lf

(x)

= X

V

it s u f f i c e s Let S

n ( f-lf(x) ) V

U

be

to p r o v e

subset

there

S

exists;

set in

is a p l a n e

contains

a

as

Cn

,

H

U and

t

X --a

~ k.

Hence

is i r r e d u c i b l e .

if

X

by the

point,

of

a

and

that

dim a S + k = dima

of d i m e n s i o n

isolated

when

neighbourhood

such

in fact, then

3

small)

of

= {a}

Snf-lf(a)

an o p e n

Proposition

a (sufficiently

an a n a l y t i c

(Such an

d i m x f -vl f v (x)

we have '

X.

is an a n a l y t i c

local

set

representation

n - k

and we may

such take

that

in theorem

Hnf-lf(a)

S = HnX.)

r

132

Let

flS = g.

Hence,

Then

by Chapter

a

is an

IV,

Proposition

analytic

set

suitably

chosen.

and

S W' = f - l f (S) nW;

let

of

W,

on

g(S).

and

dim thus,

in an o p e n

f(S W) Hence,

S~ - k for

4.

of d i m e n s i o n dimension

m,

Proof.

Suppose

a~X, U

Y

dim

such

d i m f ( a ) f(U)

that

= p,

Suppose

in

to T h e o r e m Let

h

finite

=

the

of

p = I,

for

f

b

the

unions

of

when

the result

and

subset of

f(a) so t h a t

S = dima

f(U)

X;

= f(S)

complex

complex

all

of

is o p e n

if

x~X. for all

arbitrarily at

space

space

f : X ~ Y

= m - p

are

f(a) .

locally

is o p e n .

and

x~X.

small

open

By P r o p o s i ~ o n

irreducible,

of C h a p t e r

then fibres

Y = ~

g

the

fibres

of

f.

are

where

g

in

; if

b~Y, : V ~

there. ~c C p

Corollary

3

finite. of

Hence,

at o n c e

induction

X i = { x ~ f - l ( V ) l h i ( x ) = c},

any

III,

of

is o p e n

follows by

For

an o p e n m a p

fibres

~;

13. We p r o c e e d

W c U,

f(a).

that

that

theorem

map

being

[by the r e s u l t

such

a,

>~ d i m a S,

dimensional

is a n a l y t i c Y

X

irreducible

3, t h e r e

g o f : f-i (V) ~

Proposition and

3]

(disjoint)

to p r o v e If

C p)

V

is

the p r o p o s i t i o n .

a pure

of

conversely

U

of

>~ k + d i m a

on

d i m x f-lf(x)

so that,

is a n e i g h b o u r h o o d open

a

= m - p

is a n e i g h b o u r h o o d

if

Q(fls.~) S~

a holomorphic

f(U)

is an

S' = d i m X, so t h a t W a (since X is i r r e d u c i b l e ) ' --a

of

be

by Proposition

= f(S)

g(a).

dim

a locally

that

-i

is an a n a l y t i c

2,

dim

proves

f-lf(x)

g

is a n e i g h b o u r h o o d

Hence

X

g(S)

of

Y(f(a) s

S W'

i.e.

This

Then,

if

(~

then

we h a v e

Let

and only

in

by Proposition

V.

p.

5,

point

any neighbourhood

is a n e i g h b o u r h o o d

Proposition

f(U)

be

= d i m a X.

in

sets

W

V

= g(S) nf(W)

W,

is a n a l y t i c

If

set

>~ d i m a S '

all

d i m a f-lf(S) f-lf(S)

Let

!

isolated

h

are

it s u f f i c e s

C p. from Chapter h =

III,

(h i ..... hp)

c = hl(Xo) , X o ~ f - l ( V )

I,

133

is m

a constant, - i;

Xi

and

onto

then

h (I)

an o p e n

X1

=

(h2,

set

dimxoh(1)-I but

clearly

h (I)-I

is a r b i t r a r y ,

3. E.E.

Levi's

We

give

in w h i c h of

our

4.

analytic

~(I{

subset

an

Proof.

Let

Chapter any is

IV,

as

A

X

a complex

Then

X - Y

and

hence

X ~ Y, Then,

is

by Chapter

of d i m e n s i o n

- Stein

of

in

we m a y

Bi

B,~U

g

set

of

has

(i = 1 ..... p)

A

Xo~f-l(V)

its

Proposition each

of

in

X.

many = O

on

Let is

i) , the Hence, of

a

of

have on

X - Y

X

X.

By

- 2

for

so t h a t X

X - Y

connected.

locally

irreducible

X = n.

an

analytic

closure for and

a zero

x~BnU.

on

dim

Hence,

irreducible for

an

X.

--a

is

points.

is n o w h e r e

finitely g(x)

U

we

Y

of

be m e r o m o r p h i c

P

its

case

function

X ~

8,

(Theorem

(which

a~X,

suppose

poles.

a proof

space and

~ dim

we may

Let

of

in the

Y ) A,

irreducible).

and

- p;

applied,

points

--a

(since

IV,

only

have

way

Levi

function

connected

analytic

be

any

that

Further,

set

U

for

dim

suppose

a neighbourhood in

can

singular

2 ' we have

the

is

choose

function that

X

the

any meromorphic

the

theorem

of

we

= m

since

of

E.E.

that

be

at

(p - I)

complex

a meromorphic

n - i

Remmert P

Then

also

Pe

to

to

globally

and

theorem

due

such

manifold.

map

induction,

-

example

a normal

we m a y

an o p e n

theorem

another

be

Theorem Hence

(m - i)

of d i m e n s i o n

is p r o v e d .

- Stein

X

extension

By

= h-lh(Xo ) ;

as

of

C p-I.

h (I) (Xo)

d i m _Ya 4 d i m _Xa - 2. has

in

=

theorem,

Let

is c l e a r l y

h(1) (Xo)

Remmert

following n d o m a i n s in C

dimensional,

.,hp)

contituation

the

Theorem

pure

proposition

next,

the

is

any

by

set

the

B = P point

aeY,

a holomorphic divisor)

such

components Let

ais

i - Y

134

(i = I, .... p). an integer

By the H i l b e r t

k

such

that

Nullstellensatz,

gk

is h o l o m o r p h i c

there

is

at the points

a i (i = 1 .... ,p). We assert in

U - Y.

Pf

dim Pf

IV,

nowhere ~

which

to

This

Remmert

- Stein

case

4. A n a l y t i c Let

a~X

X where

Chapter

in

clearly

that

Mk C

dimension

k-

(see e.g.

Proposition

to

5.

is an analytic

P nU.

Hence

Pf = @

Let

then

has

(by

a holomorphic in

U

(and

extension

without

using

the g e n e r a l

a direct

proof

case in the

of

f.

of

dense pl

on the c o m p l e x [i.e.

the

We have analytic

b y itself

F~

• M k be

the

The closure

k times. set of

set

is a c o m p l e x

of

Ff = -CT,Ff of

X • M k.

Ff'

in

subset.

is an analytic

Ff' c (x - P)

set of

seen,

space

X - P.

subset

the

[20]).

P = i~l Pi"

is a n o w h e r e

Let

f

functions

set of poles

{(x,fI(x) .... ,fk(x)) I x,X - P}. isomorphic

so

dependence

and the c o m p l e m e n t I.

4,

However,

be the p r o d u c t cMk

of

unless

theorem

be m e r o m o r p h i c

P

U - Y

theorem.

this

C n. long

the

in

ai~P f,

is h o l o m o r p h i c

One r e d u c e s

(fi) a ~ a ] .

IV,

Let

P.

f

is a m e r o m o r p h i c

the

to prove

and a l g e b r a i c

and

Further

component

gA

F/g k

proves

fl ..... fk

of

Proposition

Since

is rather

of poles

8).

theorem.

to that of a d o m a i n

is h o l o m o r p h i c

(B i - Y).

a zero divisor), U.

Pf

k f = g ~

is not p o s s i b l e

VI,

U.

It is p o s s i b l e

space

set

no i r r e d u c i b l e

Proposition

F

to

latter

the

by C h a p t e r

extension

function

P nU = U

contains

Now,

of

in

~ n - 2,

Chapter

the

In fact,

is c o n t a i n e d that

that

in

X x

Then

135

The proof is almost proof of C h a p t e r

identical w i t h

VI, T h e o r e m

4

that given

(page 118)

in the

and is th er ef or e

omitted. If

~

denotes

of

X x ~

on

F~

composed with

Mk,

We say that the ma p of rank

< k

fl,...,fk

(Siegel,

and

fl,...,fk

Proof.

g iv en b y

Clearly

the nat u r a l

on

point of R(f)

< k).

We say that

if there exists

If

X

is a c om pa ct

are a n a l y t i c a l l y

R(flX - P) = R(~) By T h e o r e m

subset of

a

that

~ : Ff ~ M k

of d i m e n s i o n

Mk;

which vanishes

dependent.

2 and P r o p o s i t i o n

Mk

in particular,

complex

dependent meromorphic

< k,

algebraic

implies

by

X - P.

to the t h e o r e m of C h o w

by definition,

if

such that

the C o r o l l a r y

p(z I .... ,zk) ~ O

dependent

(which,

they are also a l q e b r a i c a l l y

is an analytic

into

has a jacobian

X - P

dependent

Thimm).

projection.

in

X - P

fl ..... fk

s i m p ly that

~ O

5.

X,

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

are a n a l y t i c a l l y

p(z I ..... z k) ~ O

on

Ff

f.

are a l g e b r a i c a l l y

functions

~(Ff)

is

f l , . ~ . , fk

I, m e a n s

p(fl(x) ..... fk(x))

space,

~

at any regular

polynomial

to

then the na tu ra l m ap of

f : X - P ~ C

Proposition

Theorem

the r e s t r i c t i o n

(w 2),

there on

b ei ng

2,

< k.

~(Ff)

By

is

is a p o l y n o m i a l

~(Ff) nC k ) f(X - P). This,

fl .... 'fk

are a l g e b r a i c a l l y

dependent. Corollary

I.

dimension

If

n,

2.

meromorphic

If

algebraically

n + 1

complex

meromorphic

space of functions

on

X

dependent. X

functions

on a n o n - e m p t y

is a co mp ac t

then any

are a l g e b r a i c a l l y Corollary

X

is co mp ac t fl ..... fk

open subset of dependent.

X,

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

and the

are a n a l y t i c a l l y then

fl,...,fk

dependent are

136

This follows

at once

The above proof

is due to Remmert

It can be proved, above,

from Theorem

b y methods

see Remmert

complex [29];

also

space

to those used

functions

is an alqebraic

[2, 36,

37].

I.

[29].

similar

that the field of m e r o m o r p h i c

(irreducible)

5 and Proposition

on a compact

function field;

137

BIBLIOGRAPHICAL

NOTES

The theory of analytic sets, e s p e c i a l l y of analytic n sets in C , is d e v e l o p e d in the books of M. Herv~ [19] and

S.S. A b h y a n k a r

H. Cartan,

1953/54

C. Houzel

[12].

Rossi

[I]

and in the seminar notes of

[I0]

and 1960/61,

In less detail,

[14]. The treatment

in detail,

drawn

non-discrete

I.

in Herv~

[19]

[14]). F. Hartogs'

is proved

in

theorem, proof to

II.

due to

Theorem

J.P.

Serre.

in

[12]

stated here w i t h o u t

(and/or A b h y a n k a r theorem

[I],

stated on page

i, which is one C. }{ouzel in

are

proof

Gunning

-

3,

is proved

differentiable

and

R. Remmert,

[12]; he ascribes

remarks made

notes

apply,

theorem

- Remmert

w h i c h uses the n o r m a l i z a t i o n

theorem,

is due to

although the p r e s e n t a t i o n

the

for the the fourth, see

for

at the b e g i n n i n g

[16].

[24]. of

to this chapter. Their proof,

is difficult.

The

L. Bungart - H. Rossi

is different.

[17].

see M a l g r a n g e

above all,

Theorem 7 is due to Grauert

of the proof given here

[24];

is unpublished,

and applications,

The general

in [25].

form of the p r e p a r a t i o n

ideas of M a l g r a n g e

functions

these b i b l i o g r a p h i c a l

e.g.

The third proof we have given

theorem uses

H. Grauert

III.

(mostly a l g e b r a i c a l l y

of Houzel

For an analogue of the p r e p a r a t i o n

Chapter

is

sets over a r b i t r a r y

as stated here,

is proved by

preparation

That in A b h y a n k a r

[18].

The rank theorem, Chapter

-

in algebraic geometry.

Most of the results

are proved Rossi

work

in Gunning

although v e r y d i f f e r e n t

fields

The ideas in the treatment

from G r o t h e n d i e e k ' s

Chapter

in Herv6,

and treats analytic

c o m p l e t e l y valuated,

18-21 by

it is treated

is based on the same ideas.

quite d i f f e r e n t

closed).

Exposes

idea [7],

i38

Chapter

IV.

All known proofs of Oka's

are based on the ideas of Oka proof of the coherence

has a proof.

in

[9],

Cartan's

[26],

[9],

which

is given in

that Oka also

is not very d i f f e r e n t [27]. We have

in Grauert

- Remmert

a v e r y special case of a theorem of Grauert proofs

are, however,

given.

Another in

v e r y different

proof,

and is

[15]. These

from all these,

is

follows c l o s e l y the papers of

Bruhat - Cartan

[4, 5]

and Bruhat - W h i t n e y

[6]. The unproved results concerning in Cartan Whitney

[II] [6]

(Proposition

(Proposition

and functions, Malgrange

17,

18).

are given

the t r i a n g u l a b i l i t y of real analytic

Chapter VI.

The original

Cartan

where,

[i0]

proof of

version of this proof is given

of Grauert - Remmert

[I]

Oka

properties are given

and

in the

[27]

[21].

and K u h l m a n n see

2 given here

is given

in

A complete

[33]. Other proofs

in Rossi

it is a direct g e n e r a l i z a t i o n

in algebraic geometry.

in the book by

there is an error.

is unpublished,

The proof of T h e o r e m

sets

sets

[23].

however,

are due to A b h y a n k a r

2) and in Bruhat -

interesting metric

S. ~ o j a s i e w i c z

sets are

properties of real analytic

and applications

[24]. Further

C-analytic

15, Example

16,

M a n y v e r y interesting

[21];

[16],

from the one we have

quite different

This chapter

[ii],

paper of

followed

[12].

Chapter V. Cartan

says,

presentation.

Theorem 7 is proved

given

although Oka

and Cartan h i m s e l f

that he understands

Oka's version,

from that of Cartan,

3,

The first published

H. Cartan

to this theorem in

in a footnote

Theorem

of the ideal sheaf of an analytic

set, T h e o r e m 5, is that of refers

[26].

theorem,

The proof

[17].

is due to Kuhlmann

of c o r r e s p o n d i n g

results

139

Theorem

2 can also be p r o v e d

a l r e a d y has T h e o r e m p r o o f of T h e o r e m Chapter also

VII.

3

(or T h e o r e m

if one

5). For a n o t h e r

algebraic

2, see A b h y a n k a r

The R e m m e r t

[i0].

"geometrically"

The d e t a i l s

- Stein

are m o r e

general

form of the theorem.

We have g i v e n

in this

special

this

obtained

was

2 is due

[28,

IV, P r o p o s i t i o n

formulation

suggested

by

[20])

been proved

given

M.S.

The t h e o r e m H. K n e s e r

to R e m m e r t

E.E.

for d o m a i n s

in

in the

I, P r o p o s i t i o n

H. K n e s e r

[20].

Theorem

[29];

important

functions. Grauert

See [3].

generalization functions

a p r o o f for

cases of T h e o r e m functions

who u s e d [38].

see also

proof

o f t e n used. deduces

(just as we IV,

Theorem

7).

theorem

(see also

in

seem to h a v e

although

it has

this reason. C n)

at the end,

It can

from an a n a l o g u e functions;

were

see

on the

field

proved by

a very elementary method. The p r o o f g i v e n h e r e

[31].

applications

is due

The p r o o f b y S i e g e l

has

led

to the t h e o r y of a u t o m o r p h i c

Andreotti

Borel has o b t a i n e d

[2], A n d r e o t t i

a very

of the e a r l y w o r k of S i e g e l

as an a p p l i c a t i o n

w o r k of Borel

a more

simple

5, and the t h e o r e m

stated

in p a r t i c u l a r A.

[22]

12 for m e r o m o r p h i c

5 is due to T h i m m

to R e m m e r t to v e r y

included

37]

[15]

literature,

of C h a p t e r

[36,

with

Grauert

C n] does not

(in the case of d o m a i n s

Siegel

in the p r o o f

3 for R e m m e r t ' s

Levi

be d e d u c e d

C.L.

30];

see

Rajwade.

We h a v e

of m e r o m o r p h i c

this

5 from C h a p t e r

b e e n used.

Special

than

form m o s t

image

in T h e o r e m

[due to

in g e n e r a l

[32],

simultaneously

is the

t h e o r e m on the d i r e c t

Chapter

The

deal

case b e c a u s e

is in

complicated

given.

it from his

articles

theorem

we h a v e

Theorem

These

[i].

of the m e t h o d s

is still u n p u b l i s h e d .

-

far-reaching on m o d u l a r of

[3];

this

140

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