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 �9 Adviser: 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~

31

CHAPTER III. - LOCAL PROPERTIES OF ANALYTIC SETS

In this chapter, we will be concerned with the local

description of analytic sets, both over R and over C.

In the first section we shall deal with properties that

are valid in either case, and in the second with those

properties that are special to complex analytic sets.

A more detailed analysis of real analytic sets will be

undertaken in Chapter V. The results are mostly contained

in Remmert - Stein [32], Cartan [I0, 12], Herv~ [19].

w I. Germs of analytic sets.

Let k be either R

set in

k = C,

set in

the set

or C, and let ~ be an open

k n. Analytic functions will mean holomorphic if

real analytic if k = R. Let S be an analytic

and let a~. We denote by S the germ of --a

S at a. We refer to S as an analytic germ. --a

Let I = I(~a) denote the set of all (germs of) analytic

functions in ~ which vanish on the germ S (this n, a --a

statement has an obvious meaning). Clearly I is an ideal

in On, a"

! We have, obviously, SaCra if and only if I(~a))I(S~).

We say that S is irreducible if whenever there are two --a

analytic germs ~la' ~2a with ~a = ~laU--S2a ' one of the

germs S must be = S . --ia --a

The following lemma is obvious.

Lemma I. S --a

pr ime i d e a l .

is irreducible if and only if I(S a) is a

Since ~ is noetherian, any increasing sequence of n,a

ideals in ~ terminates. Hence any decreasing sequence n,a

of analytic germs ~la~_S2a)... terminates. We deduce easily

from this the following

32

Proposition i.

k a finite union S = LJ S

--a v=l --va

S such that, for each v, --va

Any analytic germ S can be written as --a

of irreducible analytic ~erms

S ~ S Further, this --va ~/v --~a

decomposition is uniquely determined upto order.

Definition i. The 9erms S introduced by this decomposition --v,a

S = US are called the irreducible components of S . --a -- v, a --a

Let now I be an ideal in ~ =~ ; we suppose that n n,o

{0} ~ I / 0 . If x I .. x are the coordinates of k n, n _ ' " ' n

we shall denote by O the subring of ~ consisting of p n

functions independent of Xp+ I .... ,Xn. We have a natural

injection ~ ~ ~ . Let A denote the quotien t ring ~n/I. p n

Then, we have a natural homomorphism D : ~ ~ A. P

Propow 2. After a linear change of coordinates in

there is an integer p, O ~ p < n, such that ~ : O ~ A P

is injective and makes of A a finite 0 -module �9 P

k n ,

Proof. Let f~I, f ~ O. We may make a linear transformation

of k n so as to ensure that f(O,x n) ~ O. This condition is

invariant under linear transformations ofk n-I By Chapter II �9 t

Theorem 2, (2), there is a unit u and a polynomial

qn- I

qn [ (x I = x + a .... , Xn_l) xV Pn n v n' av(O) = O, with f = UPn; O

then Pn~I" Now, either In_l = In~n_l = {O}, in which case

we take p = n- I, or there is fn_l~In_{{O~As above, we

find, after a linear change of variables in k n-l, that

there is a polynomial

qn_l -I qn-i

[ a' (x I a (O) = O, Pn_is Pn-i = x + , ) v , n-i v "'''Xn-2 Xn-l' v O

33

Continuing this process�9 we find an integer

I = In<9 = O and such that, for any r > p, P P

distinguished polynomial�9

qr -I qr [ (n-r)

= + a ...,Xr_l)Xr, a Pr Xr v (xl' v O

p such that

there is a

(n-r) (O) = O,

with P ~I = InQ . r r r

injective. If f~n'

We claim that hhis integer p satisfies our requirements.

In fact, trivially I = {0} implies that ~ : O ~ A is P P by Chapter II, Theorem 2, (i), we have

qn- I

f -- [ fl (X I .... Xn_l)X ~ (mod P ) v=O , v ' n '

q -I n-1

fl, = [ ...,Xn_2)x~ (mod Pn_l) v f2,v,~ (xl ' n-i N--O

and so on, so that

ap+ I a n f (x i ..... Xp) Xp+ 1 ...x f = [ ~ n (mod Pp+l' "'''Pn ) '

~j <qj

ap+ I an so that the images of the monomials Xp+ 1 ...x n , ~j < qj

generate A over Q P

In what follows, we shall identify elements of ~ with n their images in A =~/I when no confusion is likely�9

Corollary. If, in addition, I is a prime ideal,

quotient field of ~p, L that of A = ~n/I, then

L = K(Xp+ I ..... Xn).

K is the

Remark. The necessary and sufficient condition that the

coordinates satisfy the assertion of Proposition 2 is that

I = {O} and, for r > p, there exists a distinguished P

polynomial Qr(Xr;Xl .... ,Xr_i)~ ~r_l[Xr]nI.

34

We now state two algebraic theorems that we shall use.

I. (Theorem of primitive element). If K is a field of

characteristic zero and L = K(u I .... ,u r) a finite

algebraic extension of K, then, for any infinite subset

ScK, there exist elements cl,...,Cr~S such that

r

L = K(~) where ~ = [ c.u.. i 1

i=l

II. Let K,L be as above, and in addition, suppose that

K is the quotient field of a factorial ring A, that B

is the integral closure of A in L, and that ~EB is

such that L = K(~). Let P be the minimal polynomial

of ~ over K. (Then PEA[X] since

If P' denotes the derivative of P,

there is QEA[X] of degree < degree P

~P' (~) = Q(~) (note that P' (~) ~ O) .

A is factorial.)

then for any ~EB,

such that

Now, by the theorem of primitive element�9 there exist n

complex numbers k. such that 3

independent of x I ..... x and P

any f~ since A is a finite ~ -module, there exists n' p

m-1

a polynomial Qf(X) = X m + O [ b~(xl,...,Xp)X~,%[X] with

Qf(f) -- O. If we choose the polynomial Qf to have minimal

degree we claim that when f(O) = O, Qf is a distinguished

polynomial. In fact if not all b (O) = O, then V

m-I

X m + ~ bv(O)X~ has, at X = O, a zero of order i < m. O

By Chapter II, Theorem 2,

= L kjxj is linearly Yp+I p+1

L = K(yp+l). Further, for

where u (2) �9 Q~(X) = u.Q(x)

1-1

is a unit, and Q(X) = X 1 + [ c (x I .... ,Xp)X ~ is O

35

a distinguished polynomial of degree i. But then Q(f) = O,

and Qf would not have minimal degree. Thus we obtain (since

~p is factorial)

Proposition 3. Given a prime ideal Ir {O} / I / n' n'

there exists, after a linear change of coordinates in k n,

an integer p, O 4 p < n such that

-~A =~ : ~p n/I

is an injection which makes A a finite ~ -module. Further, P

if K is the quotient field of ~ , L that of A, we have P

L = K(Xp+l), and for any r > p, the minimal polynomial

P of x over K is in ~ [X], and is distinguished, r r p

so that there is a distinguished polynomial

qr- 1

qr [ (r) v x' Xp) Pr(Xr;X') = Xr + av (x')Xr . . . . = (xl "'"

~=O

(r) av (O') = O, with Pr(Xr,X') ~I.

It follows that if p = O, and if I = I(So), then --So is

the germ defined by S = {O}.

In what follows, we shall suppose the prime ideal

given, and the coordinates chosen so that Proposition 3

applies. We shall use the notation of Proposition 3.

Let 6 denote the discriminant of the polynomial P p+l

and OPP+I). Then ( so that 6 is the resultant of Pp+l OXp+ i

6~ , further since P is the minimal polynomial of p p+i

Xp+ i over ~ , 6 / O in ~ �9 since I = {O}, 6~I. p p' p

By the algebraic theorem II stated above, if q is the

degree of Pp+i' we have the following

Lemma 2. For any f~ there is a polynomial Rf of n'

degree ~ q - I in ~_[X] such that P

36

6f - Rf(Xp+l) EI.

In particular, there are polynomials Qr of degree ~ q - 1

in ~p[X] such that, for r > p,

6Xr - Qr(Xp+l) ~I.

Lemma 3. For any f~ there exists g~ - I and n' n

h~ such that gf - h~I. P

Proof. Since A is a finite ~-module, we have a relationship P

m-1

fm + [ a v ( fuji .Xl,...,Xp) v=O

we may suppose, since I

then only to set h = -ao,

is prime, that a O(x') ~ O.

m-1 = [ g + a (x')f u-I

v = l v

We have

Definition 2. Let S be an aqalytic set in an open set

in k n. A point agS is called a regular point of S of

dimension p if there is a neighbourhood U of a, Uc~,

such that SnU is an analytic submanifold of dimension p

of U. A point agS is called singular if it is not regular.

A point a~S is regular of dimension p if and only if

there exist functions fp+l''''' fn ~n,a such that, in a

neighbourhood of a, S = {x I fi(x) = O, i > p} and

.. are linearly independent. (dfp+l) a, -, (dfn) a

Let S be an analytic set in an open set

We suppose that ~o is irreducible, i.e. that

a prime ideal in ~ = ~ . Choose coordinates in n n,O

that Proposition 3 is satisfied. Then we have

~ck n , O~S.

I = I (S O) is

k n so

Proposition 4. There is a fundamental system of neighbourhoods

U = U'~U", U'ck p, U"ck n-p of 0 such that if ~ : (SnU) ~ U'

denotes the restriction to SnU of the projection of U onto U',

37

-I then ~ is a proper map and every fibre ~ (x'), x'~U',

of ~ is a finite set.

Proof. Choose a neighbourhood V of 0 such that the

polynomials Pr(Xr,X') of Proposition 3 all have

coefficients analytic on V and vanish on SnV. Let

= . are distinguished, ~ere V {xllxl < Q~ Since the Pr

is ~ > O, such that if Ix' I ~ ~ and Pr(Xr,X') = O,

then Ix I ~ 2/2. Clearly if U' = (x'EkPllx'l < o}, r

U" = {x"~kn-PIlx"l ~ Q), and ~ is as above, then

-I {x" x" (E) cEX I I I ~ Q/2}, so that ~ is proper. Further, -i

if x~ (x') , then Pr(Xr,X') = O, and Xr can take

at most finitely many different values.

Lemma 4. Let I be a prime ideal in ~n, and --oS the germ

of analytic set at 0 defined as the set of common zeros of

a finite system of generators of I. Let Pr' r > p,

6x r - Qr(Xp+i) be as in Proposition 3 and Lemma 2. Then

there exists a fundamental system of neighbourhoods U = U' x U"

of O such that these functions are analytic on U, S_o is

induced by an analytic set S in U, and such that the

following hold.

(a) SqUn{xl6(x') ~ O} = {x~Ul6(x') / O,Pp+l(Xp+l,x') = O,

6x r - Qr(Xp+l) = O, r > p + !}.

(b) If x~U' • k n-p and Pp+l(Xp+l,x') = O = 6x r - Qr(Xp+i),

8(x') ~ O, then x~U.

Proof. Choose V = V' x V" such that all the functions

considered are analytic on V; further�9 let f1'''" f be �9 m

analytic functions on V with SnV = {x~Vifi(x) = O,i=I ..... m}

(fi generators of I). As in the proof of Proposition 2,

we find that there exist f~,l" ~p' a = (ap+ I ..... a n), ~'3 < qj'

38

with

ap+l a n f = f . (X')X ... X (mod P ... P ) i aj <qj a, 1 p+l n p+l' ' n

and hence, if N = qp+2"''qn (substitute 6x r = Qr )

bNf i = R! (Xp+ I) (rood Pp+l " Pn 6Xp+2 -Qp+2 .... 6x - Qn ) l '" "' ' ' n '

where RI is an element in ~p[X]. Again, since Pp+1 is

monic, we may make a polymonial division of R! by x Pp+l

and obtain

6Nf i = R i(xp+ I) (mod Pp+l .... 'Pn'6Xp+2 - Qp+2 .... '6Xn - Qn ) '

where R i is a polymonial of degree r q - 1 (q = deg Pp+l).

Now fiEI; hence Ri(Xp+l) EI. Since Pp+l is the minimal

polymonial of Xp+ 1 over ~p, and deg R i < deg Pp+I' this

implies that R. = O, so that 1

6Nf i ~ O(mod Pp+l,...,Pn,6Xp+2 - Qp+2 ..... )

We now proceed as follows. Clearly, for each

we have, on V' x k n-p

bqrP r - Ar(Xp+ I) (mod bXr - Qr )

r > p + I,

where

Pp+l' we obtain

6qrP ~ A (mod 6x - Qr ) on r r(Xp+l ) Pp+l' r

where Ar~%[X ] and has degree < deg Pp+l"

Ar(Xp+l)~I and so is zero near O, hence 0

A~%[X]. Making a polynomial division of A'r by

V' x k n-p,

Again

on V' x k n-p.

39

Hence

(2) 6qrPr ~ O(mod Pp+l,6Xr - Qr ) on V' x k n-p. This

(Xp+l,x') = 6x - Q = O implies that if 8(x') ~ O, Pp+I r r '

then P (x ,x') = O. Since P is distinguished, this r r r

implies that if V' is small, then any solution x~V' x k n-p

Pp+I Xp+l' - r > p + I, lies in a of ( x') = 0 = 8x r Qr'

preassigned neighbourhood V' x V" of O. This proves (b).

Again, (I) and (2) imply that, for a fixed integer M > O,

(3) 6Mf i = O(mod Pp+l,6Xp+2 - Qp+2'''''6Xn - Qn ) o

If we now choose UcV, U = U' x U" such that (b) holds,

and all the above congruences are represented by linear

relations with coefficients analytic on U, then

snun{xi6(x') /0, f1(x) = ... = f (x) = o}, m

and, by (3) , this is

(Xp+l,x') = O = 6x - Q (Xp+ I) r > p + I}. = {x~U 16(x') ~ O, Pp+I r r '

This proves Lemma 4.

We remark that in Proposition 4 and Lemma 4, given any

~'ck n-p (which is a neighbourhood of 0), then we can find

a neighbourhood V' of 0 in k p such that for any open set

U'cV', O~U', the assertions in Proposition 4 and Lemma 4

are true.

Proposition 5. Let U be a neighbourhood of 0 such that

Lemma 4 is true relative to the ideal I = I(So) where S --o is irreducible. Then any point x~SnU with 6(x') ~ O is

a regular point of S of dimension p, and the projection

has a jacobian of rank p at x.

40

Proof. Since 6(x') ~ 0 and Pp+i

that, at the point x, @Pp+i / O. Hence, ~Xp+ i

near x by the system of equations

~r (xp+i) Pp+i(Xp+l,x') = O, x r = 6(x') '

the property that dPp+ l ..... d(x r

(x) = O, we conclude

S is defined

r > p + i,

Q~ 6 (x'))

k-independent at

which has

are

x. This proves Proposition 5.

Remark further that if X,Y are analytic manifolds

countable at ~ and f : X ~ Y is an analytic map such

that f-l(y) is discrete for every y~f(X), then

dim X { dim Y (apply the rank theorem to a point where

the differential of f has maximal rank). Combining this

with Proposition 5 we obtain

Proposition 6. The integer p of Proposition 2 and 3

_ , _ being irreducible, is the relative to I = I(S a) S a

largest integer m such that

set S), every neighbourhood of

S is regular of dimension m.

(~a is induced by an analytic

0 contains points at which

This characterisation of the integer p is clearly

invariant under analytic automorphism of a neighbourhood of

O in k n .

Definition 3. The dimension of an irreducible analytic 9erm

~a at a~k n is the integer p of Proposition 2. The dimension

_ is the maximum dimension of of an arbitrary analytic germ S a

of the irreducible components S of S �9 --v, a -- --a

an a n a l y t i c s e t S i n a n o p e n s e t ~ i..~n k n

where ~a is the 9erm at a defined by S.

The dimension of

is max dim Sa, as

41

Theorem i. Let S be an analytic set in an open set Q

in k n. Let a~S and dim S = p. Then any neighbourhood -- --a

of a contains points at which S is regular of dimension

p. In particular, the set of regular points of S is

dense in S.

Proof. --a

dimension p. Let

components of S a"

sets in an open set

T T" at a; then --a' --a

of T of dimensicn

a regular point of

that U

on T" .

to T . Then, there is --a

(since T" (T') . Let 6 --a --a

(relative to T ). Then --a

Hence, arbitrarily near

f (x) ~ O.

Let T be an irreducible component of S of --a

T' be the union of the other irreducible --a

Let T" = T nT' and T, T" be analytic -a -a -a

U' containing a inducing the germs

SnU' = TuT". Since a regular point

p which does not lie on T" is clearly

S of dimension p, it suffices to prove

contains a regular point of T of dimension p not

Let U be so chosen that Propositions3, 4, 5 apply

f~9 f = O on T" f~I = I(T a) n' --a' --

have the significance of Proposition

6~I. Since I is prime, f6~I.

a, there are points x~T with

Theorem i follows from Proposition 5.

Proposition 7. If S is an irreducible germ of dimension --a

p and S ~S', where S' is any analytic germ at a, then --a --a --a

dim S' < dim S . --a --a

_' is irreducible �9 Proof�9 Clearly, we may suppose that S a

We choose the coordinates x i, ,x n in k n �9 .. so that if

I = I(S a) I' = I(S a) (so that IcI') , then we have I = {0}, _ , p

x')~I there exists a distinguished pseudopolynomial Pr(Xr;

r > p. If we show that, after a linear change of variables

in kP(x I .... ,x ), there is a distinguished polynomial P

Pp(Xp;Xl, ..�9 I) ~I', the result follows from the remark

after Proposition 2. Now, by the preparation theorem, it

suffices to prove that there exists hs nI' = I' h ~ O. p p'

Let g~O n, g~I, g~I' By Lemma 3, there is gls - I

42

such that ggl ~ h(mod I), where h~%. But then clearly,

since gr h~I', and, since I is prime, ggl~I, so

that h~I, and in particular h / O.

w 2. Complex analytic sets.

In this section we shall deal only with complex

analytic sets, so that k = C.

= ~ , and suppose Let I be a prime ideal in ~n,o n

that the coordinates (x I ..... Xn) are so chosen that

Proposition 3, 4 and Lemma 4 are valid. Let ~ : SnU ~ U'

be the projection defined in Proposition 4.

For any ideal Ic~ we denote by S(I) the germ n,a

at a of analytic set defined as the set of common zeros

of a finite system of generators of I. Clearly, S(I) is

independent of the system of generators chosen.

Proposition 8. We have ~(SnU) = U'

Proof. Since ~ is proper, its image is closed in U'

Hence it suffices to show that ~(SnU) is dense in U'.

For any x'r 6(x') / O, the polynomial Pp+l(Xp+l,X') has

a complex zero Xp+ 1. Let

By Lemma 4, x~SnU, and clearly ~(x) = x' Hence ~(SnU~

contains the dense set {x'r ~ O).

Remarks. I. Note that, by the remark following Proposition 3,

this implies that, if the coordinates are so chosen that

Proposition 2 is valid, then there is a fundamental system

of neighbourhoods {Uu} of 0 such that ~ (SnU) is a p

neighbourhood of O in C p, ~ being the projection of C n

onto C P . P

2. Actually, the map

- s (x uI6(x') 0} u' - 16(x')

Qp+2 (Xp+l) Qn (Xp+l) x : (x',Xp+ 1, 6(x') ''''' 6(x') )"

is a covering

map.

43

Lemma 5. Let

above, for any

that ~(x) = x'

fE~ . If, for sufficiently small U as n

x'~U', 6(x') ~ O, hhere is x~SnU such

and f(x) = O, then f~I.

Proof. If f~I, there is g~I such that gf - h(mod I)

where h~ . Then h~I, and for any sufficiently small P

6(x') / 0, h(x') = f(~) g(x) = o (if x~SnU, ~(x) = x'),

so that h = O and so h~I, a contradiction.

x,

Theorem 2. (Hilbert's Nullstellensatz). Let 05 be any

ideal of ~n and --oS = S(%) the germ of analytic set

defined as the set of common zeros of a finite system of ~ m

generators of 0~ . Then I(S O ) = rad0% = {f~(Dnlf'"~ 0% for

some integer m > 0}.

Proof. We first remark that, if ~ is prime, l(So) =~.

This is a trivial consequence of Lemma 5. Hence, if ~ is

primary (i.e. rad~ is prime), we deduce, since

S(~) = S(rad~), that I(~o) = rad~. If 05 is arbitrary,

{O}, since ~ is noetherian, we obtain by the Noether n

decomposition theorem,

k = u/~--i ~v' ~u being primary.

Clearly then

k s(~) = ~I s(9),

so that

k k

u = l v = l rad ~u = rad~.

We now give a very important application of the results

obtained above. We begin with a definition.

Definition 4. Let S be an analytic set in an open set n

.in C A function f o_~n S is said to be holomorphic at

a~S if there is a neighbourhood U o_~f a i__nn Q and a

holomorphic function F in U with FIUnS = fIUnS.

44

We may define germs of holomorphic functions in the

obvious way. If a~S, let ~ denote the ring of germs S,a

of holomorphic functions at a on S. Clearly, we have

Hence S,a

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

D

is an analytic ring over C.

Definition 5. A map f : S i ~ S 2 (S i analytic set in an

open set in C ni) is called holomorphic if the map j o f,

n 2 where j : S 2 ~ C is the natural injection, has the form

jof = (fl .... f ) where the f are holomorphic on S i �9 n v "

2 Clearly, a holomorphic map f : S I ~ S 2 induces, for

a~S 1, an algebra homomorphism

f* : ~)S 2 f(a) ~) �9 Sl,a

viz, f*(9) = 9 o f-

Theorem 3.

homomorphism

Let f : S 1 ~ S 2 be holomorphic. Then the

f* : ~S 2 f(a) ~ ~ , Sl,a

is finite (see Chapter II) if and only if a

point of the fibre f-lf(a).

Proof. Let Slr162 n) , S2r162 ) .

We may suppose that a = O, f(a) = O. We set

RI = ~S1,O" R2 = ~S f(O) " 2'

Suppose that

is an isolated

f* : R 2 ~ R I

is finite. Then every element of R i is integral over R 2.

45

Hence, if e~S there exist holomorphic germs 1,0'

a l , . . . , a r ~ % s u c h t h a t 2 ' 0

r r v * [ o .

V=l

In particular, we have, in some neighbourhood of O on SI,

r r [ (k)( )<-v %

x k + a f(x) = O, k = I, .... n,a(k)~ ~=1 v v 2 , 0 "

Hence, if f(x) = O, and x is near 0 on S I, x k

satisfies a polynomial relation and so can be at most one

of finitely many complex numbers. Hence 0 is isolated

in f-lf (0) .

Suppose conversely that 0 is an isolated point of

f-lf(O). This means precisely that if ~ is the ideal

of On, 0 generated by (fl ..... fm'I(~o ))' then S(~) = {O}.

(For notation S(~), see Theorem 2). Hence, by the Null-

stellensatz, for any k, I { k { n, there is an integer r

such that

m

r [ Xk = ~v(x) f (x)(mod I(S O ) ) ~ ~

v=l v -- ' u n,o"

This implies clearly that there is an integer q > 0 such

that [~(R I) ]qcf*(~R 2)).R 1 (for notation see Chapter II).

This implies that f* : R 2 ~ R I is quasi-finite. By Theorem I,

Chapter II, f* is finite, q.e.d.

Corollary I. The necessary and sufficient condition that n

a system of coordinates (x I ..... x n) of C satisfy the

assertion of Proposition 2 relative to an ideal Ir is n

that 0 is an isolated point of the set

{X I = ... = X = O}nS(I) and In~ = {0}. P P

46

Corollary 2. If X,Y are analytic sets in open sets in

C n, ~ respectively; f : X ~ Y a holomorphic map for which

a6X is an isolated point of f-lf(a), then there is a

neighbourhood U of a such that any b~U is an isolated

point of f-lf(b).

Proof. Let X be an analytic set in an open set

C n and suppose that a = O. By Theorem 3, there exist

a ~y,b,b ~,i ~ = f(O), such that

Pi Pi-~ Pi + a f(x) x. = 0 in ~ )

xi p i i X,o p=l

in

C n. There is a

C n onto C p is a proper map with finite

then ~IS is an open map.

here Xl,...,x n are the coordinates in

neighbourhood U of 0 and an open set VcY, b~V, so

that a are holomorphic in V,f(U) cV, and the above ~,i

equations hold on U. Then, given f(x) eV, each x. can 1

have only finitely many values, which proves our assertion.

Corollary 3. If S is an analytic set in an open set ~cC n

of dimension p at every point and the restriction to S of

the projection ~ of

fibres into ~' = ~(~) ,

Proof. Let O~S and U a neighbourhood of O; we have to

show that ~(UnS) is a neighbourhood of 0 = ~(O)~C p. This

is an immediate consequence of Corollary 1 above and the

Remark 1 after Proposition 8.

Corollary 4. If S is an analytic set in an open set ~ in

C n and OeS, then dim So is the smallest integer k such

that there exists a subspace H of C n of dimension n - k

such that 0 is an isolated point of HnS.

This follows easily from Corollary I above and the

definition of dim ~o"

Corollary 5. If X is an analytic set in an open set in C n

and f : X ~ C k is a holomorphic map, then any point a~X

47

has a neighbourhood U in X such that, for x~U

f-lf(x) ,< dim f-I dimx a f (a) .

Proof. If p = dim f-lf(a), then, if the coordinates at a

a are suitably chosen, a is an isolated point of the set

f-lf(a) n{z~cnlzl = = z = O}. P

Let g : X ~ C k+p be the map g(z) = (f(z),zl,...,z ). Then -I P

a is an isolated point of g g(a) ; by Corollary 2 above,

there is an open set U' in C n containing a such that, -i

for z~UnX, x is an isolated point of g g(x). This

means precisely that x is an isolated point of

f-lflx ) n{z,cnlzl = x I ..... Zp = Xp}.

By Corollary 4 above dim f-lf(x) ~ p. x

We now continue our study of complex analytic sets.

n Theorem 4. Let ~ be an open set C and ~, the

restriction to Q of the projection of C n onto C p

(first p ~ariables). Let ~' = ~(~) and A' be a thin -I

subset of ~'. Let X be an analytic set in ~ - ~ (A')

and suppose that ~ I X is a finite covering of ~' - A'

(i.e. ~I X is proper and locally biholomorphic) and that

~I ~ .is a proper map into n' Then ~ is an analytic

set in ~ of dimension p at each of its points.

-i Proof. Let A = ~ (A'). We may suppose that ~ (and

hence ~') is connected. Hence ~' - A' is connected

(Chapter I, Proposition II) and hence there is an integer

k such that for any x's - A', there are exactly k

points x (1),...,x (k) ~X with ~(x(J)) = x'

Let f be any holomorphic function on ~. We define

holomorphic functions al, f .... ,ak, f on ~' as follows.

48

For x'c~' - A', let al, f

symmetric function

be the l-th elementary

1 ( x ' ) = ( - 1 ) (Jl)) f(xCJl > al, f [ f(x

l~Jl<...<Jl{k

(I) (k) where x ,...,x are the points of X

Clearly al, f is holomorphic in ~' - A',

: X ~ H' is proper, for any compact set K'r a I f

is bounded on K' - A', and hence (Chapter I, Proposition 10)

can be extended to a holomorphic function on Q'. Let Pf

be the holomorphic function on ~ defined by

k

Pf(x) = fk(x) + [ al,f(x')fk-l(x), x(x) = x' 1=1

with ~x(J))= x'

and further, since

By construction, Pf(x) = 0 if xEX, and hence Pf(x) = 0

if x~X . We claim that

m i

(4) X = {XCnlPf(x) = O for any holomorphic f on ~}.

Let X' be the set of common zeros of the Pf, let x'c~',

and set E '= {xcX'l~(x) = x'}, ~ = {x~l~(x) = x'}. Then

EcE', and to prove (4), it suffices to prove that E = E'

For this, it suffices to prove that f(E) = f(E') for any

holomorphic f on ~. Let ~cf(E') ; because of the

continuity of the roots of a polynomial, there is a sequence

x' of points, x' ~ x', x'cA', such that there is a zero V D V

k

of the polynomial k + [ al (x,u) ~k-I such that the v 1 'f

sequence a -~ ~; but since x' cA', there is x cX, V V D

~(x u) = X', with ~ = f(x ) . Further, since x : X ~ ~' D V V

is proper, we can find a subsequence {Uk} such that

x~X; clearly then x~E and f(x) = lim f(xvk) = ~. xv k

Hence ~f(E), and (4) is proved. Because of Corollary 2

to Theorem 5, Chapter II, X is analytic in ~. It is

clear, since ~:X ~ H' - A' is locally biholomorphic, that

49

P

X is regular of dimension p at every point of

X is dense in X, Proposition 5 implies that X

dimension p at every point.

X. Since

has

Proposition 9. Let S be an analytic set in ~ such

that S is irreducible. Let U be a neighbourhood of 0 --o

as in Proposition 4 and Lemma 4. Then the set

X = {x~SI6(x') ~ O} is dense in some neighbourhood of O

on S.

Proof. We have already remarked that the projection

~:X ~ U' - {x'~U'I6(x') / O} is a covering. Further

: S ~ U' is proper, hence so is ~ : X ~ U'. Hence, by

Theorem 4, X is an analytic set of dimension p at O.

But since XcS, Proposition 7 implies that there is a

neighbourhood V of O with XnV = SnV.

It is also possible to avoid Proposition 7 and use

directly the argument used in Theorem 4. This method leads,

in fact, to a somewhat stronger form of Proposition 9.

Of course�9 Proposition 9 is stronger than Propositions5 and 6.

If fr f~l(S O) there is gs gf ~ h(mod I(So) ) i 0 1 ~ �9

where h~ . Replacing X in the above proof by the set

{x~Slh(x')6(x') ~ O}, we deduce

Proposition 9' If S

then the set of points

neighbourhood of 0 in

is as in Proposition 9 and f~I(So) ,

{x~S(f(x) = O} is dense in some

S.

It is trivial matter to extend Proposition 9' to sets

S for which S is not necessarily irreducible (with the --o

obvious conditions on f).

Definition 6. Let S be an analytic set in an open set

in C n. A holomorphic function f on ~ is called a

universal denominator for S at a point a~S if a has

5O

a neighbourhood

if

of

S i ,

fh

on

U in ~ such that the followin9 holds:

h is a holomorphic function on the set S' of points

SnU at which S is regular, and if h is bounded on

then there is a neighbourhood V of a such that

is the restriction to S'nV of a holomorphic function

V.

Theorem 5. Le__~t S be an analytic set in ~, and suppos e

that S has dimension p at each point. Let ~ denote

the projection of C n onto C p (first p variables),

and let ~' = ~(~). Suppose that ~IS is a proper mapping { -1

into ~' with finite fibres i.e. ~ (x')nS is finite

for any x'~'}. Then, 9iven a point a~S which is regular

on S at which x I - a I .... ,x - a form a system of local -- p P

coordinates, there exists a linear function 1 on C n and

~olomorphic functions ~I .... '~k o__nn ~' such that~ if we

set

k

P(~,x') = k + [ a (x') ~k-~ v

~=I

the following conditions are satisfied.

(a)

(b)

on s.

0P l(a) ,~(a) M O, P (x) = l(x),~(x) is a

universal denominator for S at every one of its points.

(c)

(d)

If h is holomorphic and bounded on the set S' of

regular points of S there exist holomorphic functions

~o,...,~k_l on ~' such that

k-i

P' (x)h(x) = [ ~u(~(x))(l(x))~ on S', and there is ~=0

a constant M (independent of h) such that

(e) II~ II , .< MIIhlIs,.

51

Proof. We begin by proving the following. There exists -i

a thin subset A' of Q' such that if A = Sn~ (A')

then • : S - A ~ ~' - A' is a (finite) covering. Note that,

by Corollary 3 to Theorem 3, A is nowhere dense in S.

Let B be the union of the set of singular points

of S with the set of points of S' where the jacobian

matrix of the map ~ : S' ~ ~' is not invertible. (Since

: S' ~ ~' has finite fibres, this is a nowhere dense

analytic set in S'.)

Clearly, B is closed, hence so is ~(B) = A'. We

claim that A' is thin. For this, it is clearly sufficient

to prove the following: if b'~' and b~S, ~(b) = b',

then, there is a neighbourhood U of b and a thin set E'

in a neighbourhood U' of b' in ~' such that S is

regular at any point of Un S - ~ (E') , and

~iUn(S _ -I(E,)) is of maximal rank Let S be the �9 --~,b

irreducible components of S at b. By Corollary i to

Theorem 3 and Lemma 4, there is a holomorphic function g~

near b' (which is a multiple of the discriminant 6 V

corresponding to --~,S b ) such that g~I(S , b)_ , while

{x~S ,b(g~(x') = O} contains the two sets ~ (~g,bn~v,b)

and - , {xES ,bI6_ u(x') = O}. Because of Proposition 5, we may

gv ! take for E' the zeros of g' =

Thus A' is thin�9 Since ~: S - A ~ ~' - A' is

proper and a local hcmeomorphism, it is a finite covering.

We now choose a linear function l on

(i) for a (countable) dense set of points

separates the points of ~-i(x') ;

n C such that:

x's - A' , 1

(ii) for the given point a, l(a) is different from l(c) -i

for any c ~ a, c~ ~(a).

Suppose that ~ : S - A ~ ~' - A' is a covering of k

sheets (note that, by the Corollary to Theorem 3 and Proposition 8

52

: S ~ ~' is an open map). Let ~I ..... ~k be the

holomorphic functions on ~' - A' which are the elementary

symmetric functions of the values of 1 on the fibres

of ~. They have holomorphic extensions to ~' We set then

k

P(~,x') = ~k + [ a (x')~ k-v v

v=1

Clearly, P(l(x) ,~(x)) = O on S - A, hence, this latter

set being dense in S) = O on S. This is (a).

Since by assumption x I - a I ..... x - a form local P P

coordinates at a on S, ~ is a homeomorphism in a

neighbourhood of a; hence, by our assumption (ii), l(a)

is a simple root of P(~,a'), and hence

0~(l(a),~(a) ) ~ O.

To complete the proof of Theorem 5, it suffices now to

prove (c) ; to obtain the second part of (b), we have only

to apply (c) to a neighbourhood of a given point of S.

We have only to find holomorphic functions ~ on ~'

such that (e) holds and (d) holds on S - A. Let (1) (k)

x'~' - A', and let x ,...,x be the points of

with ~(x (j))- = x'. Consider the sum

S - A

this is clearly of the form

k-1

~ (x,)~v V ----O

where the ~ are holomorphic on ~' - A', and for any

~v(x') is a linear combination of the( 3)) with

coefficients depending only on the x Hence

l~v(X')~,< M max lh(x(J))l. 3

Xle

53

In particular, the

is bounded on S')

to ~' such that (e)

k p(~;x')

j=1 - [xlJll'"

are bounded on ~' - A' (since h

and so admits a holomorphic extension

holds. If in the identity

k-I

V=O

we substitute ~ = l(x) where x is a point such that

~(x) = x'~' - A' and 1 separates the points of ~-l(x')

[such points are dense in S - A], we obtain (d) on a

dense subset of S - A; hence (d) holds on S'

Remark. The set of points where S is not regular is

clearly contained in the set P' (x) = ~p (l(x),~(x)) = O.

Corollary. If S is an analytic set such that ~o is

irreducible there exists f~ - I(So) which is a n,o

universal denominator at any point sufficiently near O.

Of course, this corollary can be proved directly in a

somewhat simpler fashion. In fact, the last part of our

arguments shows that with the notation of Proposition 3, ~P

p+l is such a universal denominator. ~x

p+l We shall see later that the finiteness of the fibres

of ~ IS is a consequence of the hypothesis that ~ IS is

proper.

Remark. Theorem 4 remains valid if we replace the assumption

that ~ : X ~ ~' - A' is an unramified covering by the

assumption that ~ : X ~ ~'- A' is a proper map with finite

fibres and that X has dimension p at each point. In fact,

we have seen in the proof of Theorem 5 that there is a thin

subset B'r ~' - A' such that X - ~-I(B') is dense in X

-1 - A' B' and ~ : X - ~ (B') ~ ~' - is an unramified

covering. Then, the constuction of the functions Pf(x) can

be done, first for x'~' - A' - B', and, by successive

applications of the continuation theorem, extended first to

54

x'~' - A' then to x'~'

the same.

The rest of the proof remains

Proposition IO. (Maximum Principle) Let S be an

analytic set in ~ and f a holomorphic function on ~.

Let S be irreducible�9 and suppose that f is not --a

constant on S in any neighbourhood of a. Then f(S)

is a neighbourhood of f(a) in C.

Proof. We may suppose that a = O, f(a) = O. Let U be

a neighbourhood of O, and let the coordinates in C n be

so chosen that U = U' x U" U'c C p , , p = dim ~o and

: SnU ~ U' is proper, has finite fibres (and satisfies

Proposition 3). For x'~U', 6(x') ~ O, let al,...,a k

be the elementary symmetric functions of the values of f -I

on ~ (x') nS ; then the a admit holomorphic extensions v

to U'. Let

P(~;x') = ~k +

k av(x,) k_v

~=I

Then, for x'~U', 6(x') / O, we have

f(~-i(x') ) = {~CIP(~,x') = 0}. Hence, by the continuity

of the roots of a polynomial, we have

f(~-1(x')) = {~CIP(~,x') = O} for any x',U'.

By assumption, we have f~I(~o). Hence, by Lemma 5,

ak(x') ~ O near O~U'. Further, a (O) = 0 for each v v

(since f(O) = O). We may suppose, after a linear change of

variable in C~ that ak(O .... ,O,Xp) ~ O near x = O.

By t h e p r e p a r a t i o n t h e o r e m , t h e r e e x i s t s a d i s t i n g u i s h e d

p o l y n o m i a l

m

Q(Xp �9 1 .... ,Xp_f~) = x TMp + [ b u (x I, .... Xp_ f~)x m-vp v=l

such that P and Q have the same zeros near O. But clearly�9

55

since b (O) = O, for any ~ near O, there is x near v p

O with Q(Xp,O .... ,O,~) = O; hence the set ~C near O

for which there exists x' near O with P(~,x') = O is

a neighbourhood of O. Since f(~-i(x')) = (~CIP(~,x') = O},

f(S) is a neighbourhood of O.

Corollary i. A compact analytic set

of a finite number of points.

n S in C consists

Proof. It suffices to prove that for any holomorphic

function f on C n, f(S) is finite. If it were infinite,

there would exist ~f(S) - (f(S))o and a ~f(S) a / v ' v

if v ~ U, such that a v ~ ~. Let sv6S , f(sv) = ~v; by

passing to a subsequence if necessary, we may suppose that

= , then s v ~ SoES, since S is compact. If ~s O U_Sk,so

there are infinitely many s v on at least one ~k,s o.

Hence f is not constant on this component, so that f(S)

is a neighbourhood of f(s o) = a, and then

a~f(S) - (f(S) 1 O, a contradiction.

Corollary 2. Let ~ be a proper holomorphic map of an

analytic set S in an open set in C n into an open set -I

in C m. Then, for y~C m, ~ (y) is a finite set (being

a compact analytic subset of cn).

Corollary 3. Let ~ be an open set in C n, x : ~ ~ C p

the projection. Let ~' = ~(~) and A' be a thin subset -I

of ~' If X is an analytic set in ~ - ~ (A') of

dimension p at any point and ~I~ is proper, then

is an analytic set in ~ of dimension p at each point.

This follows from Corollary 2 and the remark preceding

Proposition 10.

Proposition II. Let S be an analytic set in ~ such that S --a

is irreducible. Then a has a fundamental system of neighbour-

hoods U such that the set of regular points of S in U is

connected.

56

Proof. Choose U such that Proposition 4 is valid and

further, with the notation as before, the set

Sn{x~U'!6(x') ~ Oj is dense in SnU. It suffices to prove

that X = Sn{x~U!6(x') / O} is connected. Now

: X ~ U' - {x'~U' 1 6(x') = O} is a finite covering. Hence

if X is not connected, and Y is a connected component

of X, ~I Y is also a covering. Let h be the functbn = O

on Y, = i on X - Y. If S' is the set of regular points

of S, S' - X is thin, and hence h has a holomorphic

extension to S' By the corollary to Theorem 5, there is

f~I(S_o) such that F = fh is holomorphic on ~o" Now,

for any x'~U', 6(x') ~ O, clearly there is xs

(viz x~Y) with F(x) = O. By Lemma 5, Fs so that

F vanishes at all points near O. But clearly there are

points arbitrarily close to O on X - Y where f / O,

which implies that F I(So). This proves Proposition II.

Corollary. If S, a are as above, and

there is a neighbourhood V of such that

at any point of VnS.

dim S = p, then --a

S has dimension p

This follows at once from Proposition ii and Proposition 5.

Theorem 6. Let

in C n and let

Then there exists a nei~hbourhood

S be an analytic set in an open set

O~S. Suppose that S is irreducible. --o

V o_~f O and finitely

fl "'" f in V such that: ' ' m

many holomorphic functions

(a)

(b)

Proof. Choose the coordinates (xl,...,x n) in

U' = {x',cPi,x ' , < e'}, U" = {x"~Cn-Pl.x", e"} <

if U = U' • U", then the projection

proper with finite fibres.

the set of singular points of SnV is precisely the

set {x~Vlfl(x) = ... = f (x) = O}- m

each f is a universal denominator at every point l

of V.

n C and

such that

: UnS -~ U' is

57

Let e > O be sufficiently small, and set, for I sij I < ~,

i = i .... ,p, j = i ..... n,

n

i j=l ~ij j

Then, if e is small, ~i ''''' ' Xp+l .... 'Xn

linearly independent. Let Qi < Q' be fixed, and e

are

small

�9 ' = {x'~U'l Ix' I < Qi } and U I ' x U" enough Let U I = U i .

I (a) (x) i < QI } and let ~ : SnU ~ ' Let Ua = {x~U I~ i , ~ ~ U I

be the map ~a(x) = (~)(x) ,...,~p(~) (x)). If e is small

enough, then ~ is proper. To prove this, we remark that if

', (K') K' is a compact subset of U I ~s is closed in U,

since clearly it cannot be adherent to any point of ~U

in U and is closed in U . Further, if ~ (x)~K', then

' in U if e ~(x) lies in a compact neighbourhood of U I -i

is small enough; hence ~s (K') is contained in a compact

subset of U, and being closed in U, is itself compact. -I

Because of Corollary I to Proposition Ii, ~ (x') is S

finite for any x'~U I.

Now, for any regular point a of S in U, there

exist ~ij' laijl < e, such that ~s)" ,.-.,~)- form a

system of coordinates at a. Let W be a neighbourhood of

U s (such a W exists if ~ is O such that We l~iJ 1 <~

small). If a~WnS and S is regular at a, let a be so (~)

chosen that the ~ form coordinates at a. There exists,

by Theorem 5, a function P' which is a universal denominator s at any point of Us, P'~(a) / O, which vanishes on the

singular set of S in U s. Hence, there exists a family

of holomorphic functions {ft} in W which are universal

denominators at any point of W such that the singular set

of S in W is given by {x~Vlft(x) = OUt}. Theorem 6

follows easily from this and Chapter II, Theorem 5 (if we

replace W by a smaller open set V).

58

Corollary I. Let S be an analytic set in an open set n

~2 in C Then the set of singular points of S is an

analytic set in ~.

proof. Let a~S and S = US where the S are --a --~,a' --~,a

the irreducible components of _S a. Let Su be an

analytic set in a neighbourhood W of a inducing S o ~v,a

Since the S are irreducible and none of them is con- B~,a

tained in the union of the others,we have

dim (Su, nS a ) < min (dim S a' dim S a ) if N ~ u. Hence, -- a --~, --p, --~,

by the Corollary to Proposition II, we may choose W so

small that for b~S nS nW we have u

dim (S ,bnS , b)_ < min (dim S ,b' dim S b ) for ~ / u.

In particular no germ S is contained in the union of ' --u,b

the others. We claim that the set of singular points of S

which lie on S is the union T of the set of singular P U

points of S with ~ (SvnS ~) . In fact, it is clear

that S is regular at any point of S v not on T u. If

bs , u ~ ~, then since an analytic set is clearly

irreducible at regular point, b is singular. If

b4 ~u (S~nSu) then S b = S so that b is singular

on S if and only if it is singular on S u. The corollary

clearly follows from this and Theorem 6.

If S is not irreducible and if S = U S is --O --O B~,O

the decomposition of ~o into irreducible components,

let S, S be representatives of these germs in a neighbour-

hood of o. If f is a universal denominator for S u,b

for any b~S nU, and if g is holomorphic on U, g = O U

on S , N / v, g / 0 on a dense subset of S u, then h = gf

is a universal denominator at any point of SnU for S.

It follows from this that we have

Corollary 2. Theorem 6 remains valid if we drop the

hypothesis that ~o is irreducible.

59

Corollary 3. If S is an analytic set in an open set

Q in C n, and A is the set of singular points of S,

and if aES, then for any f~n which vanishes on A ,a --a'

there is an integer k ) I such~ that fk is a universal

for S at all points near a.

This follows at once from Corollary 2 above and the

Hilbert Nullstellensatz. n

Proposition 12. Let ~ be an open set in C and Ac~

an analytic subset of dimension ( n - 2. Then, for any

holomorphic function on ~ - A, there is a unique holomorphic

function F on ~ with FI~ - A = f.

Proof. The uniqueness of F, if it exists, is obvious.

Hence we have only to prove that for every a~A, there is

a neighbourhood V and an F holomorphic in V with

FIV - A = flV - A. We may suppose that a = O, and, after n

a linear change of coordinates in C , that O is an

isolated point of the set

An{xlx i : .. :x --0). �9 n-2

If Q > O is sufficiently small, and

i 0), a~ = (O,...,O,~,

it is clear that, if ~ is large, the set

and

-D = I ~x~C n x I =...=x u n-2

I = O, x =--, IXnl ~ Q}c~- A, n-I

K o = {x~cnlxl = = Xn_ 2 = O, Xn_ 1 = O, IXn l= Q}cn - A.

Hence, by Chapter I, Proposition 12, There is a connected

neighbourhood V of O (containing Ko) and F holomorphic

in V such that F = f near K o. Since V - A is

connected, it follows from the principle of analytic

continuation that F = f in V - A.

60

Proposition 13.

in C n , and let

for a~S, f ~ O

S induced by S --a

we have

Let S be an analytic in an open set

f be holomorphic in ~. Suppose that,

on any irreducible component of the germ

at a. Then, if S' = {xs = O},

dim S' = dim S - I. --a --a

Proof. i. We first suppose that S = ~ and a = O. Then,

by the preparation theorem, we may suppose that

P

f(x) = xP +n [ a~(x')xP-U,n x' = (xl, .... Xn_1) , a (O) = O. v=i

n-i Clearly, if U = U' • U" U' U" , c C , c C, is a neighbour-

hood of 0 such that x'~U' f(x' x n , , ) = 0 imply that

~U" then the projection ~ : S' = {x~Ulf(x) = O} ~ U' x n

is surjective; hence if g = g(x I ..... Xn_l) ~I(So) , then

g =- O, {since g clearly vanishes on ~(S')}, so that, by

definition of dimension, dim S' = n - i. D O

2. For the general case, we may suppose that a = O

and that --oS is irreducible �9 If p = dim So, we can find

a neighbourhood U = U' x U" of O, U'r C p U"c C n-p such

that ~ : SnU ~ U' is surjective, proper, and has finite

fibres�9 Further, if 6 is as in Lemma 2, W = {xs ~ O}

is dense in SnU. Now ~ : W ~ U' - {x'~U'16(x') = O} is a

finite covering, of say, q sheets Let, for x'r

6(x') ~ O, h(x') be the product of the values of f at

the points of S lying over x' Then h is bounded (if

U is small enough) and so admits a holomorphic extension,

also denoted by h, to U' Clearly {x'~U', 6(x') / O,

h(x') = O~ is the projection of WnS' onto U'. Further, J

if x~S, 6(x') = O, we can find a sequence of points

xv~S, 6(x' v) ~ O, x v ~ x. If f(x) = O, then f(x ) ~ O,

hence h(x' v) ~ 0 so that h(x') = O. Hence

61

x(S') r = 0}, and one proves, in the same way,

the converse inclusion , so that x(S') = (x'EU'[h(x') = 0}.

Since f ~ 0 on S h ~ O. By the argument given in I. ~0 �9

above, we can suppose, after a linear change of coordinates

in C p (and shrinking of U') that the following holds:

let

of

Xp_ i be the restriction to U' of the projection

C p onto C p-i Then T = x (x(S')) = x (U') �9 p-I p-1 "

It follows since any function g (xl, .... Xp_l) ~I (_So)

vanish on T, that I(S_~) n~D_ 1 = {0}, so that

dimS =p- i =dimS o- •

Theorem 5 has another important application�9

must

Theorem 7. Let S be a m analytic set in an ope n set in C n.

The space of holomorphic functions on S is complete; i.e. if

(fp} is a sequence of holomorphic functions on S and

f ~ f, uniformly on compact subsets of S, then f is p

holomorphic on S.

Proof�9 Let aeS and S v be analytic sets in an open k

neighbourhood ~ of a such that --aS = 9=kJl S a is the --9 t

d e c o m p o s i t i o n o f S i n t o i r r e d u c i b l e c o m p o n e n t s . By --a

Theorem 5, applied to S after a linear change of co- --9,a

ordinates in C n , t h e r e i s a n e i g h b o u r h o o d V 9 o f a i n

and a holomorphic function u in V such that if 9 V

is holomorphic on V nS, there is a holomorphic function U

~ in V 9 such that�9 with M' > 0 independent of ~,

we have {x~V nS u (x) fi O} is dense in V nS 9 9 9 V 9'

Uv~ = ~ in VgnS9,

and

ti~vilvv ~ M' ii~11VvnS.

If gv i s a h o l o m o r p h i c f u n c t i o n i n V v a n i s h i n g on 9

~/g S~ ,

62

while {x~Selge(x) ~ O} is dense in SenV e (such a ge

e x i s t s , b y P r o p o s i t i o n 9 ' i f V i s s m a l l e n o u g h ) , t h e n , e

for any ~ holomorphic on V uS, there is a holomorphic e

~e (= ge~) on V with

(a) v ~ = ~e on V uS, e e

(b) ll~ellve ~ MII~IIVenS ;

= and M > O is independent of @. Further, here v e geue

{x~Senvelve(x) ~ O} is dense in SenV e-

be the ring of germs of holomorpHic

a, and ~a = 5a (S), the ideal in

Let ~a =~R,a

functions on ~ at

of functions ~vanishing on S a.

Consider the homomorphism

~ : ~a~% k

given by ~(f) = (vlf .. vkf) Let E a = a(%) + ~kr '' ' " a a"

By Cartan's theorem (Chapter II, Corollary 1 to Theorem 5),

if a sequence (hp) of k-tuples of holomorphic functions

in a neighbourhood U of a converges uniformly on U

(to h say) and (hp) a~Ea for each p, then (h) a~Ea.

Let {f } be a sequence of holomorphic functions on P

which converges uniformly on ~nS. Let ~o = fo'

~p = f - f p ~ I. We may suppose that p p-l'

co

ll~pllQn S < co. p=o

functions ~p (e = 1 ..... k) on V ,e V

V enS, ll~p,eIIv e ~ MII~IlvenS" If on

then (a(~p))a- (~p) a~ k, hence (0p) a~E a. Further,

converges on V = NVe since [ ll~pllv ~ M[ ll~pll~n S ~ co.

By our remarks above, there are holomorphic

such that Ve~p = ~p, e

~p = (~p,l ..... ~p,k )'

63

Let ~ = (~(I) ,~(k) ) = [ @p Then, by our remark above i o �9 �9 " I

(@)a~Ea, so that there is a neighbourhood W of a and (v)

a holomorphic function ~ on W such that vv~ =

on WnS.

Let f = lim fp = [ ~p. Then, since Vv~p = @p,v'

we have vvf = ~)on WnS, so that vv(f - ~) = O on WnS

for ~ = I .... ,k. But since {x~VvnSvlvv(x) # O} is dense

in V nS, this implies that f = ~ on WnS~ for each ~,

so that ~ = f on WnS. Since @ is holomorphic on W,

this proves the theorem.

Remark. The idea of this proof is essentially that of

Bungart-Rossi [7].