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~ AI,
(8)
Thus,
we have o nl y to prove
is c e r t a i n l y
c 2 > A 2.
Clearly,
valid
we have,
m
(lo) i=l
for
~+u
v = O if
g~R',
(8). The if
that
14
and,
further,
(11)
' ICy+qi(g) l ,
Icy(a i(g))l
where
qi = (0 .... ,q ..... O)
(q
in the
i-th place).
Hence we have
(12)
I c~(bj (g))I
"< mA 1 ~+y=~
A2l~Imax i
Ic
(g) I" Y+qi
Assume now that
(8)
is valid for a certain
and apply
to
g = b . . . . b. (f), j = 91. 32 3~+ 1
(12)
and all
(X0
We obtain
(13)
l%(bjl "''bj~+l (f))I
"< reAl
A21~fClC21yl+qc3 ~
~+y=c~
(c~)'~'} ~ ClC2
= ClC 2 lal c 3 V{mAlC2 q
I~l
c3
v+l
if
(14)
c 3 > ~1c2 q
(A2~I
,~,
(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
h(O)
1
/ O,
this
implies
~ O.
h(O)
(2).
P = xp + n be c a l l e d
a u (O)x pn - ~ ;
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
remark
P b
u
in
remark
o f the
(2) are
then
a x P-~ u n
form
u=l
au
We
relation
(I) ; w e
= af +
as f o l l o w s .
.x p - ~ n
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,
the
exactly
p
multiplicity).
= a f + 1
for
Suppose
p '- vx b ~ n
are holomorphic v Then we have
Vx, I
in
<
E,
3
P (a -
a') f =
(b'
this
implies
P (b' I
polynomial p
zeros
in
follows
that
Remark.
It
the
k
to
case prove
that
tx
n
for
v
fixed
x
;
x ~l .... ' X n - l '
- bv)xP-~ n
I < 6;
v
I
of
this
degree
implies
fx.l 3
< e,
~ p -
I
b
= b'
that
v
the
has
at and
v
least it
a = a' is
sufficient
= R
follows
(I).
If we
to prove from
this;
have P
~=af+
bx p-~ ~n
Theorem
2 when
k
in
we
only
fact
have
= C;
18
where b'
9,
f~R{x},
= Re(by) ,
v
a, b
~C{x}, t h e n v a', b' ~R{x}
then
, if
a' = R e ( a ) ,
and clearly
v
P
= a'f + V
In the n e x t we
shall
Theorem
two proofs
therefore 2'.
neiqhbourhood
f
be
< r I ....
is z e r o
that
the o r i g i n
that
f (x) / 0
only
'
fx
for
is a z e r o for
ho lomorphic
b I , . . . ,b p
of the f < r
n
x
n
a
a
= O,
and a holomorphic
zero
such
holomor~hic
(16)
p.
n
I ~ r
in
that ,
and
m o r e co v e r
Suppose
fx n f -- r n .
on
Ix 3.I
< r j,
Ix.l 3
< r., 3
there
j ~ n
and
there f
< rj,
such
j ~ n - i}
on
u
function
P.
al, . . . , a p
functions
that
on
P
on which
P' is
that
f = u(xP
and
fx
P ~ b x p-~ n v=l
exist
P
in a
suppose
j ~ n - I,
functions
there
on
in
function
Also,
Moreover,
and
}
= 0
Ixj I ~ rj,
~ = a f +
nowhere
n
function
o__nn
(15)
(0)
2,
set
of o r d e r
P' = { ( x i , . . . , X n _ I) ~ c n - i I ,xjl
v
for T h e o r e m
k = C
a holomorphic
.Then, for a n y h o l o m o r p h i c exist
that
of the c l o s u r e
P = {x~cnl .Xl, f ( O , x n)
that we give
suppose
Let
b' x p-~. v n
1
exists such
~ a .xP-V). n u=l
a constant
that,
lla,p
+
)lib
in
M
(15),
It p,
> O, we have
~ Mli~lip.
depending
only
19
Remark.
Before giving
germ of h o l o m o r p h i c a linear change numbers
r
satisfied Proof.
function
at
of c o o r d i n a t e s
> O 3 on
the proof,
we note that for any
o~C n, n in C ,
there
exist,
arbitrarily
such that the conditions
after small
of T h e o r e m
2' are
P.
We begin by proving
the existence
u, a
of
V
Let
x' =
(x 1,...,xn_ 1) ~P', 1 ~o (x') = 2xi
Then,
~o(X')
Ixnl
( rn,
of
~o(O) x';
< r n.
a k(x')
is the a~
~-th
a
k
in a continuous
>~ O, 1 2~i
"
on
P'.
t k Of dt -~(x',t) f(x,,t ), Itl=r
n
If
tjl(x ')...tj(x~ ')
(-I) u l~Jl <...ju~ p
elementary
is a polynomial
so that
for
is h o l o m o r p h i c
a (x') = v
f(x',x n)
Further, ~o(X') is c l e a r l y
+ (tp(X') }k _ "
Sk(X')
n
~ (x') = p for x'(P'. Let o the zeros of f(x',t) in
denote
Then we have,
Of dt -~(x',t) f(x',t) "
of zeros of
= p.
= (t l(x') }k +.
so that
Itl=r
hence
tl(x'),...,tp(X') Itl
/
is the number
and
function
and set
in
symmetric ~1' .... ~p
is h o l o m o r p h i c
in
function
of the
(with rational
P'
t., then 3 coefficients),
Let
v
P ~(x',Xn)
= xPn +
a~(x')x p-n u. u=l
Then,
for any
x'~P',
f(x' ,X n) ~(x,x n)
1 2~i
f
and
~
/
have
the same zeros.
f(x' ,t)
Itl=r
~(x',t) n
dt t-
xn
Hence
2O
~(x''Xn)
_
I
f ( x ' , x n)
/
2~i
~(x',t)
ftf=r
dt
f(x',t)
t-
xn
n are holomorphic
P.
on
Hence
P.
and nowhere
zero
replace
P
by
altering
the c o n d i t i o n s
are b o u n d e d
on
P,
I 2~i
Clearly,
the
so t h a t
a
and
f
is h o l o m o r p h i c
=
of
by on
without u
coefficients a n d the
dt t - Xn
-1
and
of
U
~.
inequality
remarks
(16),
above).
where
P.
in
a ( x ' , x n ) ~ ( x ' , x n)
O
< r
If
< r
'
is i n d e p e n d e n t
is h o l o m o r p h i c
@ ( x ' , x n)
that we may
Hence
of o u r
is h o l o m o r p h i c
let
~(x' l t) ~(x',t)
integral
the
(in v i e w
P,
/ ItI=r
2'.
algorithm
~
u
polycylinder
Theorem
so are
where
it is c l e a r
larger
the d i v i s i o n
replace
a(x,,Xn)
Further,
a slightly
To p r o v e we may
on
f = u~,
r
of Then,
1 = 2~i
. n
if
Ix f < r, n Ix I < r, n
if
/ ~(x',t){1Itl=r
~(x''Xn) dt x ~(x',t) } t - "
P 2~i
Itl=r
f
~(x',t)
I
/ Iti=r
~(x' ,t) ~(x',t)
- 2~i
where
c
(x',t)
tP
xpn
a v(x') (t p - V - xP-~)}n t dt'x
+
n
~=I
P ~,=1
c ~ ( x ' , t ) . x Pn- ~ d t ,
is a p o l y n o m i a l
in
t
< p
of degree
whose
V
coefficients
are
b
(x')
linear
=
we
obtain
1
- a
ItI=r
= 1
on Ix.l 3
P
and ~ r., 3
of
/ ~(x' 't)
2~i
U
combinations
~(x'
b x p- ~. v n
,
t) c
the
a
e
If w e
(X' ,t) dt, V
Moreover,
the
a
are b o u n d e d V
(x' ,t)
is b o u n d e d
j ( n - I,
Itl = r
n
away .
IlbvIIp, 4 MIf~IIp,
set
V
from
Hence
0
on the
set
n
21
where
M
depends b ux p-~ n
- a~ =
o n l y the
aV
we d e d u c e
t
~.
e
Hence,
that
since
for fixed
x'~P'
V
we
have
M' II cpllp la(x',Xn) I It
n
inf f=Ix
l~(x',t)
n
I
'
i
so t h a t lim la(x',Xn) I 4 M"ii~llp. Ix l~r n n By the m a x i m u m
principle,
this
implies
that
llalIp ~ Mil~llp.
We n o w g i v e using
a third proof
ideas of M a l g r a n g e Let
f
by
x'
in the v a r i a b l e s 6
> 0
are c h o s e n
2. We d e n o t e
as above.
(x,t)
t = that
2 when
k = C,
24.
be as in T h e o r e m
(x i .... ,Xn_ i)
of T h e o r e m
Consider
n
by
x,
and
the p o l y n o m i a l
p-I ~ t xv v=O
-- x p +
(to ,...,tp_ I) ,x. if
x
Suppose
it f ( ~, ~(x,t)
that
e
> O,
has e x a c t l y
p
V
zeros
in
D = {Ixl
holomorphic We a s s e r t
< 6},
and s u p p o s e
in the d o m a i n
that there
that
~ = {x'~U',
exist holomorphic
x~D,
To p r o v e
in ~L' ={x'~U',itl p-i ~ = A~ + B oX ~ V ~=0
this w e p r o c e e d
as follows.
Itl
Expand
O~U'
A(x,x',t)
< e},
~
is
< e}, w h e r e
fulnctions
and B v ( x ' , t ) (~= O , . . . , p - l ) (17)
~(x,x',t)
in
such that
in a p o w e r
series
22
CO
(18)
~(x,x't)
=
~v (x',t) x ~, v--O
where
~v
We wri t e
is h o l o m o r p h i c
on
this
~ = Ck + ~ , k where
=
~k
~v
"X V
,
~
=
v=O as
k ~ co,
on compact
is a m o n i c p o l y n o m i a l ; on
coefficients functi o n s
B
~
(19)
on
~')
in
k.
a sequence
Bv,kr ~ B y
as
then o b t a i n
B
r ~ co.
the zeros of < e},
in
x
with
.k.X ~-
on c o m p a c t
Once this a s s e r t i o n
Since
of integers ~k ~ ~
subsets
is proved,
such that
as
k ~ co,
we
Akr ~ A, we w o u l d
(17).
for fixed
{Itl
Now,
V
are b o u n d e d
{kr}
We start by p r o v i n g Let,
~.
and h o l o m o r p h i c
+
{~,Bv,k}
uniformly
can choose
~ O
. p-l) such that '"" ' p-1
~k = ~ x
~,
~
there exist
v=O
of
Clearly
(which is a p o l y n o m i a l
(v = O
We assert that
.
subsets of
hence,
holomorphic v,k
V
v >k
uniformly
holomorphic
~v.x
t, ~;
the
Jtf if
xi(t)
the b o u n d e d n e s s
of the
< e, Xo(t) ..... Xp_l(t) t
lies in a c om pa ct lie in a c o m p a c t
{By,k}" d en ot e
subset of
subset of
D.
We have p-1 (20)
~k(Xj (t),x', t ) :
Let ~;
6
6(t)
denote
is a p o l y n o m i a l
(x',t) xj
B v=O
the d i s c r i m i n a n t in the
t
of the p o l y n o m i a l
which V
if we c o n s i d e r
(t) iv.
~,k
the s q u a r e d d e t e r m i n a n t
is
~ O.
F ur th er
23
det
(xj (t)U)o ~ j, u { P _
this is a holomorphic only when
6
does
function of
t,
i 2 which vanishes
(it is in fact a constant multiple
of
6). Hence I det(xj(t) u) I is bounded away from zero on compact {6(t) ~ 0}.
It follows
from
(20)
~'.
Chapter
However, I
since
Theorem
This,
and
imply that
uniformly bounded on compact
{~}
= O).
{~}
Let now variables
f
for any
by Theorem
aF(x,x',t ) , b~,F(x',t)
at
i, Chapter
subsets of
2. Introduce
and the polynomial
F~C{x,x'},
is
~.
~
the new as above.
there exist holomorphic O,
germs
such that p-i
(2i)
F(x,x')
= aF.~ +
b F
x~
~=O In particular p-• (22)
f(x,x')
We claim that
= a(x,x~t)~(x,t)
a(O) ~ O;
in fact,
+
~ b (x',t)x ~ ~--O
(22)
gives
I,
This,
(17).
be as in Theorem
t o .... ,tp_ I
{Ak}
subsets of Again,
proves
is uniformly
~; hence
is uniformly bounded on compact
as remarked earlier,
Then,
~'.
subsets oz
n - {(x,x',t) I ~(x,t)
6(t) ~ O
~,k
subsets of
(19),
bounded on compact
is
B
'
bounded on compact
B
is holomorphic on ~', ~,k implies that B is uniformly
i
'
that
~,k subsets of the set
uniformly bounded on compact in
subsets of the set
24
p-1 (23)
f(x,O)
b
= a(x,O,O).x p +
(O,O)x v, V
v=O
which,
since
f
has a zero of o r d e r
p
at
O,
implies
that
(24)
a(O)
# O,
b
(0,0)
= O.
V
We n o w c l a i m
T. (x')
near
that
O ~ C n-I
there
exist h o l o m o r p h i c
T. (0) = 0
such that
3 b
functions
and
3
(X',T(X'))
~ O.
TO p r o v e
this,
because
of the r a n k
V
theorem,
it
suffices
to prove
that
the matrix
0b
( -a-V (o,o)) is n o n - s i n g u l a r . respect 0
to
t
Now,
(22)
if we d i f f e r e n t i a t e
and put
x' = O,
with
we obtain,
t = O,
for
~ ~ < p,
(25)
O = a(x,O,O)x ~ +
Since
~,u
< p,
this
Oa X p
(x,O,O)
i~ttl
implies
p-I r Ob ~ ~ (O,O) x ~. v--O 0t
+
~b 0t ~ (0,0)
that
0b v < ~,
if
~b
P (0,0)=~)tp
is a t r i a n g u l a r so that
= 0
a(O,O,O)/
O.
matrix with non-zero
Thus
('~l~j ( 0 , 0 ) )
diagonal
elements,
it is n o n - s i n g u l a r .
Thus
there
a neighbourhood such that
b
exist h o l o m o r p h i c of
functions
O ~ C n-I 0 ~ j ~ p - I,
(X',T(X'))
= O.
Substituting
T. (X') , 3 Tj (0) = O, in
in
(22) , we h a v e
V
p-I (26)
f(x,x')
= a(X,X',T(X')){X p +
=0
Tj (X')XJ}~
J here
the p o l y n o m i a l
in b r a c k e t s
has
the p r o p e r t y
required
25
in T h e o r e m in
2, w h i l e
(21),
we
a (O) / O,
obtain,
for
and
is a u n i t .
= aF(x,x',T(x')){x
jcx xJ}§
p + j=O
which,
with
Theorem
(26)
the
fact
p-1 u=O
that
a(O)
~"
/ O,
proves
for
Theorem
2.
The is d u e
fourth
and
hast
to H . G r a u e r t
Suppose has
and
Substituting
any p-i
F(x,x')
a
a zero
and
that
of
proof R.
p
we
Remmert.
f~k{x},
order
that
and
at
x
;
then
Here
that
k
= R
or
2
C.
f ( O , . . . , O , x n)
= O.
n
give
~ O,
Write
co P
f =
f ~ ( x i .... ,Xn_i) x n
0
for
v
< p,
f
divide
f
suppose
that
We
by
suppose
Let
by hypothesis
fv(O)
that
= O
if
for
f = =
v
<
a ~ x a, ~s
p,
series we
il~ll =
a Banach
Ic
space
of p o s i t i v e
have If
~ =
~I an Ir i ...r n
relative terms
6,
may
this
ri =
< Q,
...
= r
= 6
r)
for
u
<
p,
norm.
-1.
< co
Ir I a
..r "
< co n
on~
and
Further,
since
in
any way
we
like,
~s
n-i
Ilfelt ~:M16 (where
ic
a norm
be m u l t i p l i e d ~,
f = O f v(Xi,...,Xn_i) x n V
of
to
fat
laai Q
c xa a ' is
tl~ti ~ it,ll li~ti if
GO
v < p,
= O
fp(Xl, .... Xn_ 1)
then
i' " " " 'rn
is
(0) V
(0) ~ O. C l e a r l y , for T h e o r e m 2, w e m a y P a n y u n i t in k { x I ..... X n _ l } . Hence we may
=
Clearly,
f
r
= r
< Q,
and
n we have,
M1,...
since
fv(O)
are constants
= 0
for
independent
and
IL fxl .....XnlXnll M2rP+I
26
Hence
IIf _ x pll = llf _ fp(X 1 .... ,Xn_l) xP i ~ M36
Given
~
> O,
6 = 6(r)
< Q
If
e < 1,
we can c h o o s e
such that
~E~,
M36
r
+ M2rP+l
< Q,
+ M2rP+l. and then
< er P,
so that
write = a(~)x p + b(~), n
where
a(~),
b(~) ~ ~
of d e g r e e
< p;
A
be the
:~
~
~
and
then
b(~)
li~ll = lla(~) ur p +
linear
~11= I l a ( ~ ) ( f -
IIA~
so that,
if
I
IIA - Ill < 1. surjective.
denotes Hence
Hence,
if
this
We shall follow the
identity ~
is invertible, ~,
there
implies
now collect
given
contained
in H e r v ~
Let of germs
a~cn(Rn),
already
and
with
(1).
various
preparation 2,2'.
let
results
theorem
isomorphism
(k = C
or
R).
Theorem
3.
~n,a
~
in the case
~n,a~---k{xl
denote
(which are
is a f a c t o r i a l simply
a = O,
at
on
n
and
a.
we h a v e
- a I ..... x n - a n }
ring.
induction
the ring
functions
Then we have
uses
that
in one of
We omit proofs
=~ a n,a (real analytic)
remarked
a canonical
The p r o o f
~
17).
of h o l o m o r p h i c
As we have
we have
in p a r t i c u l a r
is
2,
together
in T h e o r e m
~,
+ b(~).
Theorem
from the W e i e r s t r a s s
forms
n
Let
x P ) II ~ e l l a ( ~ ) IIr p ~ e l l ~ l l ,
= A~ = a(~)f But clearly,
,b(~)II.
x
+ b(~).
the
A
in
operator
A~ = a(~)f Then
is a p o l y n o m i a l
27
a reduction factorial the
to the t h e o r e m of Gauss
ring,
same way,
Theorem
4.
so is the p o l y n o m i a l
using Hilbert's
~
k = C, a m u c h
(This is due to H. C a r t a n
~
O
Let
O~C n
(the p o w e r
,o there
Go
5.
exists
basis
q
8;
ring
is a
AX.
In
we m a y p r o v e
stronger
theorem
see also H e r v 6
denoting
is true.
19).
M
be a s u b m o d u l e of o d i r e c t sum q-times). Then
a s y s t e m of ~ e n e r a t o r s
and a f u n d a m e n t a l
A
ring.
and let
I
if
theorem,
is a n o e t h e r i a n
n,a
In the case
Theorem
that
{fl ..... fp}
s y s t e m of n e i g h b o u r h o o d s
of
M O ove____~r
{Uu}
of
such that the
f. are d e f i n e d b y h o l o m o r p h i c f u n c t i o n s 3 and such that the f o l l o w i n g c o n d i t i o n s are satisfied.
U
on m
V
(a)
~
If
functions belongs
h =
(h I .... ,hq)
on
U
is a
q-tuple
such that the g e r m
to the m o d u l e
M o,
of h o l o m o r p h i c
(h)o
then there
of
h
at
0
exist h o l o m o r p h i c O
functions
(b) a --
~l,...,a
There
q-tuple
then
if
with
exists
on --
P
Up
a constant
llhllU
h =
Corollary
< ~,
there
ajfj 1.
a sequence
then
O
h =
M
> 0
of holomorphic functions exist
i~aj11U~ ~ M '
of
with
If
o
(hk)o~M ~
on --
such that
U
V
if
and
U
v
h
.
is
(h) o~Mo,
holomorphic
on
U u
lih,uv "
is a s u b m o d u l e
of h o l o m o r p h i c
with
~. 3
V
M
on
~ ~.f. j=l 3 3
functions
and if
of ~ q
n,o
and
{hk}
on a n e i g h b o u r h o o d
hk ~ h
uniformly
on
U
U,
(ho) ~M . o
Corollary
2.
Let
functions
in an open
~
be an a r b i t r a r y f a m i l y of h o l o m o r p h i c n set Qr Then the set
/~ r x ~ I f ( x ) f~ ~ t is a n a l y t i c
in
~.
= O}
28
We
state
necessary Chapter
a proposition
due
to W e i e r s t r a s s
for the s t u d y of m e r o m o r p h i c
IV).
which
functions
is
(see
The p r o o f in the case of Cn is in H e r v 6 n n R is e a s i l y r e d u c e d to that of C
The case of
elements
~ a is factorial, we m a y speak of c o p r i m e in ~ a (i.e. e l e m e n t s w h i c h have no c o m m o n
divisors
e x c e p t units).
19.
Since
Proposition
I.
If
are h o l o m o r p h i c a~,
in
is an o p e n
(analytic)
the g e r m s
sufficiently
~
fa' ga
near
a,
functions
on
are c o p r i m e
in
the g e r m s
and
f, g
such that
for an
, then for a are c o p r i m e
fb' gb
b
~b" We end this c h a p t e r b y showing
Theorem
4 can be u s e d
Let and
to give
A = k{x}/6%
u : A-~ B
suppose be
cn(R n)
set
that
and
another
i.e.
such that its image
be the ideal
in
Then clearly
there
in
k{x,y}
B
is
k{x}
k{x,y}
then
~oQ = u.
Let
by
Let
~
and
and the n a t u r a l Thus,
with
47
x i -~i(y).
~ B of the i n c l u s i o n
projection
a change
into
of notation,
k{x,y}/~, we h a v e
the following.
OU be an ideal
u
let
is an i s o m o r p h i s m
in
k{x,y}
and
u
k{x} -~ k{x,y}/0t. if
rings
~i(y) ~k{y} and
mapping
Then,
i.
We m a y c l e a r l y
= u(x i),
is the c o m p o s i t e
if
o n l y to p r o v e
(O) .
generated
: k{x,y}/~ such that,
be a n a l y t i c
morphism. 0t =
2 and
p r o o f of T h e o r e m
B = k{y}/~
a quasi-finite A = k{x~
how Theorem
is q u a s i - f i n i t e ,
it is finite.
the n a t u r a l
29
Proof. Y' =
Let
Y =
(Yl ..... Ym-i )
by induction u'
(Yl ..... Ym ) =
on
and
m,
let
that
: k { x } - ~ k{x,y'}/,JL'
Since
u
where
0%' = k { x , y ' } n 0 L
if the n a t u r a l
there
We assume,
map
is q u a s i - f i n i t e ,
is q u a s i - f i n i t e ,
a. ~k{x,y} I
(Y''Ym)
then,
it is finite.
is an integer
p
> 0
and
such that n ~m -
ai(x'y) xi~05 i=l
By T h e o r e m
f~k{x,y},
2, for any
there
exist
such that
bo, " "" ibp-l~k{x'Y' }
n
pv
f-
is a m u l t i p l e
bv(x,y')Y m
of
ymp -
a i(x,y) x.,l i=l
v =O so that this d i f f e r e n c e generate
k{x,y}/~
over
The i n c l u s i o n Further, then
u = i.u'
that
u'
if
induction
The
fact that
u'
u'
R, R'
such that
icu'
respectively,
R' and
such that
Artin-Rees
lemma,
to p r o v e
the proof.
Let
u'
results
rings
S,
Then
R.R'.
u'
k-vector
is finite
from
R'
be a
the i n c l u s i o n
be a h o m o m o r p h i s m is q u a s i - f i n i t e . ideals
We h a v e
(~' q.S) nR' r i
let
such that
: R ~ R'
is q u a s i - f i n i t e
at once
and
be the m a x i m a l
dimensional
since
it s u f f i c e s
this c o m p l e t e s
R'' S let ~ =
an i n t e g e r
k{x} ~ k { x , y ' } / ~ ' ,
that u'
is q u a s i - f i n i t e .
is a finite
is finite.
shall prove b e l o w
ring
is finite.
Let
is finite,
be a n a l y t i c
of the a n a l y t i c
i : R' ~ S
R' /~'
We
p-I
1'Ym'''''Ym
so that:
is q u a s i - f i n i t e
Lemma.
Proof.
i
hypothesis,
lemma.
subring
k{x,y'},
Thus
is the n a t u r a l m a p
Since
the f o l l o w i n g Let
in 0Z .
i : k{x,y'}/0%' -~ k{x,y}/0L
is finite.
our
lies
q and
of
R, R',
to show that
space. exists, R',
S
Let
q
by the are
be
S
by
3O
noetherian by Theorem 4.
It is sufficient to verify
that R'/(~'q. S) nR' is a finite dimensional k-vector space. But this is a subspace of S/~'qs. Since ~'qs~~ qR,S and iou is quasi-finite,
S/~'qs
is a finite-dimensional k-vector
space, and the lemma is proved.
31
CHAPTER
III. - L O C A L P R O P E R T I E S
In this chapter, description In the
of a n a l y t i c
sets,
undertaken
32,
Cartan
k
local
and o v e r
analytic
analytic
in C h a p t e r V. The r e s u l t s
w I. G e r m s of a n a l y t i c Let
to c o m p l e x
a n a l y s i s of real
in R e m m e r t - S t e i n
R
and in the s e c o n d w i t h
t h a t are s p e c i a l
A more detailed
both over
shall d e a l w i t h p r o p e r t i e s
in e i t h e r case,
properties
SETS
w e w i l l be c o n c e r n e d w i t h the
f i r s t s e c t i o n we
are v a l i d
OF A N A L Y T I C
that
those sets.
sets w i l l be
are m o s t l y
contained
12,
Herv~
19.
and let
~
I0,
sets.
be e i t h e r
R
or
C,
be an o p e n
set in
k n.
Analytic
functions will mean holomorphic
k = C,
real
analytic
if
set in
and let
the set Let
S
at
I = I(~a)
functions
k = R.
S
be an a n a l y t i c
We d e n o t e b y
We refer
denote
Let
if
to
S --a the set of all
S the g e r m of --a as an a n a l y t i c germ. (germs of)
analytic
w h i c h v a n i s h on the g e r m S (this n, a --a s t a t e m e n t h a s an o b v i o u s m e a n i n g ) . C l e a r l y I is an ideal in
in
a~.
a.
C.
~
O n , a" W e have,
We
say that
analytic germs
S --a
Lemma
~la' ~2a
m u s t be
following
I.
S --a
S a C r a!
is i r r e d u c i b l e
germs
S --ia The
obviously,
with
if and o n l y if if w h e n e v e r ~a
I(~a))I(S~).
there
= ~laU--S2a '
are two
one of the
= S . --a l e m m a is o b v i o u s .
is i r r e d u c i b l e
if and o n l y if
I(S a)
is a
prime ideal. Since ideals
in
~ ~
n,a
n,a of a n a l y t i c g e r m s
is n o e t h e r i a n , terminates.
sequence of
Hence any decreasing
~la~_S2a)...
f r o m this the f o l l o w i n g
any increasing
terminates.
sequence
We d e d u c e
easily
32
i.
Any
analytic
germ
union
S = --a
k LJ S v=l --va
of
Proposition a finite S --va
such
that,
decomposition
Definition
{0} ~
each
The
are
now
be w r i t t e n
as
analytic
~erms
Further,
this
~ S ~/v --~a
determined
upto
order.
9erms
S introduced by this decomposition --v,a the i r r e d u c i b l e c o m p o n e n t s of S . --a
called
I
can
irreducible
S --va
v,
is u n i q u e l y
i.
= US S --a -- v, a Let
for
S --a
be
an
ideal
in
~
n
=~
n,o
;
we
suppose
that
I / 0
. If xI .. x are the c o o r d i n a t e s of k n, n _ '" ' n we shall denote by O the s u b r i n g of ~ consisting of p n functions independent of X p + I .... ,Xn. We h a v e a n a t u r a l injection Then,
we
~
~ ~ . Let A d e n o t e the q u o t i e n t r i n g p n have a natural homomorphism D : ~ ~ A. P
Propow
2.
there
is an
After
integer
a linear
p,
O
~ p
change
< n,
of c o o r d i n a t e s
such
that
~
~n/I.
kn ,
in
: O
~ A P
is
injective
and makes
of
A
a finite
0
-module 9 P
Proof. of
Let
kn
so as
invariant Theorem
Pn
=
then we
x
f~I,
qn n
linear
(2), t h e r e qn- I
+
Pn~I"
take
to e n s u r e
under 2,
f ~ O.
a
O Now,
p = n-
that
after
a linear
there
is a p o l y n o m i a l qn_l -I qn-i = x n-i + O
a linear
transformations is a u n i t
either I,
make
f ( O , x n) ~
u
v (x I .... ,Xn_l) xVn'
find,
Pn-i
We may
or
In_l there
change
of
O.
9
= O,
= {O},
fn_l~In_{{O~As
variables
v a'v (x I , " ' ' ' X n - 2 ) Xn-l'
condition
is
By C h a p t e r
II
t
a polynomial
av(O)
= In~n_l is
This
ofk n-I and
transformation
in
with
in w h i c h above,
k n-l,
a v, (O)
f = UPn;
= O,
case we
that
Pn_is
33
Continuing
this process 9
p
we find an integer
I
= In<9 = O and such that, P P d i s t i n g u i s h e d polynomial 9
for any
such that there
r > p,
is a
qr -I Pr
=
qr Xr
(n-r) +
a
(xl' ...,Xr_l)Xr,
v
av
(n-r) (O) = O,
O with
P ~I = I n Q . r r r We claim that hhis integer
In fact,
trivially
injective.
If
I
= {0}
P
p
implies
b y Cha pt er
f~n'
satisfies that
our re qu ir em en ts .
~ :O
II, T h e o r e m
2,
~ A is P (i), we have
qn- I f --
fl
v=O q
n-1
,v
(X I .... X n _ l ) X ~ '
(mod P ) n '
-I
fl, v =
f2,v,~ (xl ' ...,Xn_2)x~n-i
(mod Pn_l)
N--O and so on,
f
so that ap+ I an f ~ (x i ..... Xp) Xp+ 1 ...x n
=
(mod Pp+l' "'''Pn ) '
~j
so that the images of the m o n o m i a l s generate
A
In what their
images
Corollary. quotient
over
< qj
Q
P follows, we shall in
~j
A =~/I
el em en ts
w h e n no c o n f u s i o n
If, in addition,
field of ~p,
identify
L
I
that of
is a prime A
= ~n/I,
of ~ n is likely 9
ideal,
K
w it h
is the
then
L = K( X p + I ..... Xn). Remark.
The n e c e s s a r y
coordinates I
and su ff ic ie nt
s a t i s f y the a s s e r t i o n
condition
of P r o p o s i t i o n
that the 2 is that
= {O} and, for r > p, there exists a d i s t i n g u i s h e d P polynomial Q r ( X r ; X l .... ,Xr_i)~ ~ r _ l X r n I .
34
We n o w I.
state
(Theorem
of p r i m i t i v e
characteristic algebraic ScK,
two a l g e b r a i c
zero
there
exist
element).
and
extension
theorems If
that we
K
is a field
L = K(u I .... ,u r)
of
K,
then,
elements
shall of
a finite
for any infinite
cl,...,Cr~S
use.
such
subset
that
r L = K(~)
where
~ =
c.u.. i
i=l II. Let K
K,L
be as above,
is the q u o t i e n t
is the
integral
such that of
~
If
P'
closure
L = K(~).
over
there
K.
denotes is
A
Let
P
numbers
k. 3
suppose
ring
L,
A,
and that
be the m i n i m a l A
since
the d e r i v a t i v e
of
that ~EB
B is
polynomial
for any
such
P
that
is factorial.)
then
P,
< degree
(note that
of
in
PEAX
b y the t h e o r e m
independent
in addition,
of a f a c t o r i a l
of d e g r e e
~P' (~) = Q(~)
complex
and
of
(Then
QEAX
Now,
field
1
~EB,
that
P' (~) ~ O) . of p r i m i t i v e
such
that
x I ..... x
and
element 9 n =
Yp+I
L p+1
there
kjxj
L = K(yp+l).
exist
is l i n e a r l y
Further,
for
P any
f~
n'
since
a polynomial Qf(f) degree
-- O.
A
Qf(X)
is a finite m-1 = Xm +
If we choose
we c l a i m
O
~
p
-module,
f(O)
= O,
Qf
Qf
with
to have m i n i m a l
is a d i s t i n g u i s h e d
p o l y n o m i a l . In fact if not all b (O) = O, then V m-I Xm + ~ b v ( O ) X ~ has, at X = O, a zero of order O By C h a p t e r
II, T h e o r e m
is a unit,
and
Q(X)
2,
exists
b~(xl,...,Xp)X~,%X
the p o l y n o m i a l
that w h e n
there
(2) 9 1-1
= X1 +
Q~(X) c
O
= u.Q(x)
(x I .... ,Xp)X ~
i < m.
where is
u
35
a distinguished and Qf ~p
would
polynomial
of d e g r e e
not have m i n i m a l
degree.
But then
Q(f)
Thus we o b t a i n
= O,
(since
is factorial)
Proposition there
3.
exists,
an integer
Given after
a prime
a linear
p, O 4 p
< n
ideal
change
such
is an i n j e c t i o n K
which makes
is the q u o t i e n t
A
field
Ir
{O} / I / n' of c o o r d i n a t e s in
n' k n,
that = ~ n/I
: ~p -~A
if
i.
a finite
of
~
, L
~
-module. P that of A,
Further, we have
P L = K(Xp+l),
and for any
r
> p,
the m i n i m a l
polynomial
P
of x over K is in ~ X, and is d i s t i n g u i s h e d , r r p so that there is a d i s t i n g u i s h e d p o l y n o m i a l qr- 1 qr = Xr +
Pr(Xr;X')
(r) v av (x')Xr .
x' Xp) . = . (xl. "'"
~=O (r) av (O')
= O,
It follows
with
that
if
the g e r m d e f i n e d In w h a t given,
We
Let
6
( so that p Xp+ i
,
Lemma degree
by
shall
we
shall
over
of 2.
For
P
p+i 6 / O in
theorem
Pp+i'
we h a v e
any
~ q - I
f~ in
then
n' ~_X P
the prime
so that
--So
is
of
and
Pp+l
since
II stated the
3
3.
of the p o l y n o m i a l
P p+l
OPP+I). Then OXp+ i
is the m i n i m a l ~ p'9
ideal
Proposition
of P r o p o s i t i o n
the d i s c r i m i n a n t
since
~p,
I = I(So),
suppose
chosen
is the r e s u l t a n t
further
if
use the n o t a t i o n
denote 6
and
S = {O}.
follows,
By the a l g e b r a i c degree
p = O,
and the c o o r d i n a t e s
applies.
6~
Pr(Xr,X') ~I.
polynomial
I p = {O},
above,
if
q
of
6~I. is the
following
there
is a p o l y n o m i a l
such that
Rf
of
36
6f - Rf(Xp+l) EI. In p a r t i c u l a r , in
~pX
there
are p o l y n o m i a l s
such that,
for
r
Qr
of d e g r e e
~
q - 1
> p,
6Xr - Qr(Xp+l) ~I.
Lemma
3.
For
any
h~
such that
f~
there
n' gf - h~I.
exists
g~
n
- I
and
P Proof.
Since A is a finite
~-module, P
we have
a relationship
m-1 fm +
a v .Xl,...,Xp) ( fuji v=O
we m a y
suppose,
since
then o n l y to set
I
is prime,
h = -ao,
=
g
v=l
Definition in k n.
2.
Let
A point
dimension
p
such that
SnU
of
U.
agS
agS
a~S
We
S
suppose
a prime
in
~o
point
are
U
linearly
of
of d i m e n s i o n
p
i > 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
if
f
ap+l . (X')X a, 1 p+l
N = qp+2"''qn
... X
a n
(mod P
n
(substitute
p+l'
... P ) ' n
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
~pX.
monic,
we m a y make a polymonial
Again,
since
division of
R!x
Pp+1 by
is
Pp+l
and obtain
6Nf i = R i(xp+ I) (mod Pp+l .... 'Pn'6Xp+2 - Qp+2 .... '6Xn - Qn ) ' where Now
Ri
is a polymonial
fiEI;
polymonial
hence of
implies that
of degree
Ri(Xp+l) EI.
Xp+ 1
over
R. = O, 1
~p,
r q - 1
Since and
Pp+l
this
so that
We now proceed as follows. on
is the minimal
deg R i < deg Pp+I'
6Nf i ~ O(mod P p + l , . . . , P n , 6 X p + 2
we have,
(q = deg Pp+l).
- Qp+2 . . . . ).
Clearly,
for each
r > p + I,
V' x k n-p bqrP r - Ar(Xp+ I) (mod bXr - Q r )
where
A~%X.
Making a polynomial
d i v i s i o n of
A'r
by
Pp+l' we obtain 6qrP r ~ A r(Xp+l ) (mod Pp+l' 6x r - Qr ) where
Ar~%X
Ar(Xp+l)~I
and has degree and so is zero near
on
< deg Pp+l" O,
hence
V' x k n-p, Again
0
on
V' x k n-p.
39
Hence (2)
6qrPr
implies then
P
implies
that r
~ O(mod if
8(x')
(x ,x') = O. r that if V'
Pp+I (Xp+l' x')
of
preassigned Again, (3)
Pp+l,6Xr ~ O,
P
is small,
V'
Pp+l,6Xp+2 UcV,
U = U'
and all the a b o v e c o n g r u e n c e s relations
with coefficients
by
(3) , this
r
of
Lemma
We r e m a r k ~'ck n-p
U'cV',
x U"
/0,
f1(x)
in a
integer - Qn )
U, =
M
by
(b).
> O,
o
(b)
are r e p r e s e n t e d on
x k n-p
This p r o v e s
such that
analytic
x~V'
lies
- Qp+2'''''6Xn
holds,
linear
then = f
...
m
(x)
= o},
= O = 6x r - Q r (Xp+ I) 'r
> 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
*
In particular,
o.
V=l
we have,
in some n e i g h b o u r h o o d
of
O
on
SI,
r r xk +
Hence,
a
~=1
if
satisfies
(k)( v
f(x)
f(x)
)<-v
= O,
= O,
and
a polynomial
x
k = I, .... n,a(k)~ v
is near
relation
0
on
2,0"
S I, x k
and so can be at m os t one
of f i n i t e l y m a n y c o m p l e x numbers. in
%
Hence
0
is i s o l a t e d
f-lf (0) . Suppose
conversely
that
0
f-lf(O).
This means p r e c i s e l y
of
generated
On, 0
(For n o t a t i o n stellensatz,
by
S(~), for any
is an i s o l a t e d that if
~
is the ideal
(fl ..... fm'I(~o ))' see T h e o r e m
then
2). Hence,
k, I { k { n,
point of
there
S(~)
= {O}.
by the Null-
is an integer
r
such that m
r Xk =
This implies that This
v=l
~v(x) f (x)(mod I(S O ) ) ~ ~ v -' u
c l e a r l y that there
~(R I) q c f * ( ~ R implies
Chapter
II,
Corollary
that f*
I.
2)).R 1
is finite,
is an integer
(for n o t a t i o n
f* : R 2 ~ R I
and s u f f i c i e n t
(x I ..... x n)
assertion
2 relative
0
of
P
= O}nS(I)
such II).
By T h e o r e m
and
In~
P
C
condition n
= {0}.
that
s a t i s f y the
to an ideal
is an i s o l a t e d point of the set
{X I = ... = X
> 0
see C h a p t e r
is q u a s i - f i n i t e .
a syste m of c o o r d i n a t e s
that
q
q.e.d.
The n e c e s s a r y
of P r o p o s i t i o n
n,o"
Ir
n
is
I,
46
Corollary C n,
~
a6X
2.
If
X,Y
respectively;
are a n a l y t i c f : X ~ Y
is an i s o l a t e d p o i n t of
neighbourhood p o i n t of
of
a
in o p e n sets
a holomorphic
f-lf(a),
map
for w h i c h
then there
such that any
b~U
in
is a
is an i s o l a t e d
f-lf(b).
Proof. Cn
U
sets
Let
X
b e an a n a l y t i c
and suppose
a ~,i ~~ y , b , b
that
a = O.
= f(O),
set in an o p e n set
By Theorem
in
3, t h e r e e x i s t
such that Pi
Pi + xi here
Xl,...,x n
neighbourhood that
a p=l
f(x)
p i
Pi-~ x. = 0 i
are the c o o r d i n a t e s U
of
0
in
C n.
in
a n d an o p e n set
VcY,
~
X,o
)
There
is a
b~V,
so
a
are h o l o m o r p h i c in V,f(U) cV, a n d the a b o v e ~,i e q u a t i o n s h o l d on U. Then, g i v e n f(x) eV, e a c h x. can 1 have only finitely many values, which proves our assertion. Corollary
3.
If
of dimension
p
the p r o j e c t i o n fibres
into
Proof.
Cn
O~S
Cn
and
onto then
U
and
If
OeS,
such that This definition
Corollary
0
~IS
S
then
of
5.
f : X ~ Ck
to
S
of finite
is an o p e n map.
of
of C o r o l l a r y
of
O;
we have
0 = ~ ( O ) ~ C p. 1 above
to
This
and the
8.
is an a n a l y t i c dim So
a subspace
set in an o p e n set
is the s m a l l e s t H
of
Cn
is an i s o l a t e d p o i n t of
follows
~cC n
is a p r o p e r m a p w i t h
is a n e i g h b o u r h o o d
1 after Proposition 4.
Cp
a neighbourhood
consequence
that there exists
and
of
~(UnS)
Corollary
set in an o p e n set
at e v e r y p o i n t a n d the r e s t r i c t i o n ~
is an i m m e d i a t e Remark
is an a n a l y t i c
~' = ~(~) ,
Let
show that
S
integer
k
of d i m e n s i o n
~
in
such n - k
HnS.
easily from Corollary
I a b o v e a n d the
dim ~o"
If
X
is an a n a l y t i c
is a h o l o m o r p h i c
set in an o p e n
map,
set in
then any point
a~X
Cn
47
has
a neighbourhood
U
in
a
such that,
f-lf(x)
dimx
Proof.
X
for
x~U
,< dim a f-I f (a) .
If
p = dim a f-lf(a), then, if the c o o r d i n a t e s at s u i t a b l y chosen, a is an i s o l a t e d point of the set
are
f-lf(a) n{z~cnlzl
=
=
z
= O}.
P Let a
g
: X ~ C k+p
is an i s o l a t e d
there for
is an open z~UnX,
means
x
be the m a p point set
of
U'
g
g(z) -I
in
that
x
g(a) ;
4 above
We n o w c o n t i n u e Theorem
4.
Let
restriction (first
p
subset
of
to
~'.
~I X
~I ~ set in
~
Proof. hence
~')
(Chapter k
such
points
Let X
~ IX
map
into
A = ~
-i
for a n y
holomorphic
p
~,
Cn
onto
and
A'
the Cp
be a thin -I ~ - ~ (A')
set in of
~' - A'
biholomorphic) ~
and that
is an analytic
We m a y suppose
II)
~' - A'
and h e n c e - A',
with
be any h o l o m o r p h i c functions
and
covering
Then
Hence
x's
sets.
at each of its points.
(A').
x (1),...,x (k) ~X f
n'
n
of
~' = ~(~)
and l o c a l l y
I, P r o p o s i t i o n
Let
C
is a finite
is connected.
that
set
be an a n a l y t i c
of d i m e n s i o n Let
of
= x I ..... Zp = Xp}.
of the p r o j e c t i o n
is p r o p e r
.is a proper
2 above,
dim
be an open
Let
that
point
f-lf(x) ~ p. x s t u d y of c o m p l e x a n a l y t i c
~ Q
by Corollary
is an i s o l a t e d
~ariables).
and suppose (i.e.
our
Then
containing a such that, -i point of g g(x). This
f-lflx ) n{z,cnlzl By C o r o l l a r y
). P
Cn
is an isolated
precisely
= (f(z),zl,...,z
that
(and
is c o n n e c t e d
there
there
~
is an integer
are e x a c t l y
k
~(x(J)) = x' function
al, f .... ,ak, f
on
on ~'
~.
We d e f i n e
as follows.
48
For
x'c~'
symmetric
- A',
let
be the
al, f
l-th e l e m e n t a r y
function
al, f ( x ' )
=
(Jl)) f(xCJl>
1
(-1)
f(x l~Jl<...<Jl{k
where
(I)
x
Clearly
,...,x
al, f
: X ~
H'
(k)
are the points
is h o l o m o r p h i c
is proper,
is b o u n d e d
on
and hence
to a h o l o m o r p h i c
be the h o l o m o r p h i c
function
and
set
~
~x(J))=
since
aI f
I, P r o p o s i t i o n
on
defined
x'
further,
K'r
(Chapter
function
on
with
X
~' - A',
for a n y c o m p a c t
K' - A',
can be e x t e n d e d
in
of
Q'.
Let
x(x)
= x'
10)
Pf
by
k Pf(x)
= fk(x)
+
al,f(x')fk-l(x), 1=1
By construction, if
x~X
.
We c l a i m
m
(4)
if
xEX,
and h e n c e
Pf(x)
= 0
i
X'
= O
for any h o l o m o r p h i c
be the set of c o m m o n
and set EcE',
= 0
that
X = {XCnlPf(x)
Let
E '= {xcX'l~(x) and to prove
For this,
f
continuity
on
~.
of points,
of the
Let
that
f(E)
D
let
= x'}.
x'c~',
E = E' for any
of the
there
such that
~}.
Then
that
= f(E')
~cf(E') ; b e c a u s e
x'cA',
on
Pf,
to prove
of a polynomial,
x' ~ x',
f
= {x~l~(x)
it suffices
to prove
of the roots
V
zeros
= x'}, ~
(4),
it suffices
holomorphic
x'
Pf(x)
is a sequence
there
is a zero
V
k of the p o l y n o m i a l
k
+
al 1
v sequence
a
-~ ~;
but
since
x' cA',
V
with
~
D
xv k Hence
clear,
we can
5, C h a p t e r
since
there
is
x
cX,
Further,
since
x : X ~
~'
V
then
and
the
D
find a s u b s e q u e n c e
clearly
~f(E),
to T h e o r e m
= f(x ) . V
x~X;
such that
V
~(x u) = X', is proper,
(x,u)~k-I 'f
(4)
x~E
and
is proved.
II,
~:X ~ H' - A'
X
{Uk} f(x)
such
= lim f(xvk)
Because
is a n a l y t i c is l o c a l l y
that
in
= ~.
of C o r o l l a r y ~.
2
It is
biholomorphic,
that
49
P
X
is r e g u l a r
X
is d e n s e
dimension
of d i m e n s i o n
in p
Proposition
X,
p
at e v e r y p o i n t
Proposition
9.
Let
S
be an a n a l y t i c
S is irreducible. Let --o as in P r o p o s i t i o n 4 and L e m m a X = {x~SI6(x')
~
We h a v e
~:X ~ U' -
O}
U
X
has
set in
~
such
be a n e i g h b o u r h o o d
of
0
of
O
Hence,
by
4. T h e n the set
is d e n s e
Theorem
already
{x'~U'I6(x')
: S ~ U'
in some n e i g h b o u r h o o d
4,
remarked / O}
is proper,
But since
hence
XcS,
Proposition of
O
directly
the a r g u m e n t
in fact,
to a s o m e w h a t
If where
fr h~
i01
.
Replacing
{x~Slh(x')6(x')
Proposition
~ O},
9'
If
of
0
is
{x~S(f(x)
C n.
universal
Let
in
S
A holomorphic denominator
9.
than P r o p o s i t i o n s 5 gs
and 6.
gf ~ h ( m o d
= O}
I(So) )
9 and
is d e n s e
f~I(So) ,
in some
S.
to e x t e n d
S is not n e c e s s a r i l y --o c o n d i t i o n s on f).
6.
leads,
in the a b o v e p r o o f b y the set
for w h i c h
Definition
7 and use
4. This m e t h o d
is as in P r o p o s i t i o n
S
It is t r i v i a l m a t t e r
obvious
O.
is a
form of P r o p o s i t i o n
there X
at
we deduce
then the set of p o i n t s neighbourhood
in T h e o r e m
9
there
Proposition
9 is s t r o n g e r
f~l(S O) ~
that
p
XnV = SnV.
stronger
Proposition
~ : X ~ U'.
7 implies
to a v o i d
used
Further
set of d i m e n s i o n
with
It is also p o s s i b l e
Of course 9
so is
is an a n a l y t i c
V
that the p r o j e c t i o n
is a covering.
X
neighbourhood
in
that
S.
Proof.
S
Since
at e v e r y point.
that
on
5 implies
X.
of
Proposition irreducible
be an a n a l y t i c function for
S
f
9' to sets (with the
set in an o p e n
on
at a p o i n t
~
is c a l l e d a~S
if
a
set a has
5O
a neighbourhood if
h
of
SnU
U
in
~
is a h o l o m o r p h i c at w h i c h
S
then there
fh
is the r e s t r i c t i o n
that
5.
and if
is a n e i g h b o u r h o o d
S
Le__~t S
to
V
S'nV
S'
h
of
of p o i n t s
is b o u n d e d a
on
such that
of a h o l o m o r p h i c
let
into
of
for a n y
with
p
Cn
~' = ~(~).
~'
S
be an a n a l y t i c
has d i m e n s i o n
the p r o j e c t i o n
on
on the set
holds:
function
V.
Theorem
and
function
is regular,
S i,
on
such that the f o l l o w i n 9
x'~'}.
Cp
Suppose
9iven
x I - a I .... ,x
- a p
coordinates, ~olomorphic
there e x i s t s functions
and s u p p o s e ~
denote
variables),
is a p r o p e r m a p p i n g
-1 ~
a point
--
p
~IS
{ i.e.
~, Let
(first
that
fibres
Then,
at w h i c h
at each point.
onto
finite
set in
(x')nS
a~S
is finite
which
is r e g u l a r
form a s y s t e m of local P
a linear
~I .... '~k
function o__nn ~'
1
on
such that~
Cn
and
if we
set k P(~,x')
= k
+
a
v
(x') ~k-~
~=I the
following
conditions
(a)
(b)
on 0P l(a) ,~(a) universal
(c)
are satisfied.
If
h
regular
M O,
P
denominator
is h o l o m o r p h i c points
~o,...,~k_l
of
on
S ~'
s. (x) = for
l(x),~(x) S
at e v e r y one of its points.
and b o u n d e d there
is a
on the
set
exist holomorphic
S'
of
functions
such that
k-i (d)
P' (x)h(x)
=
~u(~(x))(l(x))~
on
S',
and there
~=0 a constant (e)
M
(independent
II~ II , .< MIIhlIs,.
of
h)
such that
is
51
Proof.
We b e g i n b y p r o v i n g
a thin then
subset
of
• : S - A ~
by Corollary Let of
A'
S
B
the
has
Clearly, claim that
~ : S' ~ finite
There exists -i A = Sn~ (A')
if
(finite)
A
covering.
is n o w h e r e
B
is closed, is thin.
S'
where
this
hence
U'
if
and
U
b'
of
in
components
3 and L e m m a
of
S
b
rank
4, t h e r e
~(b)
,
set
S
E'
is
and S
b e the
--~,b
By Corollary
is a h o l o m o r p h i c
(which is a m u l t i p l e
= b',
and a t h i n
Let
b.
sufficient
such t h a t
(E')
at
= A'. W e
b~S,
~'
Un S - ~
is of m a x i m a l
~(B)
9
b'
(Since dense
it is c l e a r l y
b'~'
of
at a n y p o i n t of
~iUn(S _ -I(E,))
near
S.
the j a c o b i a n
is a n o w h e r e
so is
For this,
is a n e i g h b o u r h o o d
in a n e i g h b o u r h o o d
Theorem
in
is n o t i n v e r t i b l e .
fibres,
the f o l l o w i n g :
irreducible
dense
that,
set of s i n g u l a r p o i n t s
of
~'
Note
S'.)
A'
there
regular
3,
set of p o i n t s
set in
to p r o v e
is a
b e the u n i o n of the
~'
analytic
then,
~' - A'
of the m a p
: S' ~
such t h a t
3 to T h e o r e m
with
matrix
Q'
the f o l l o w i n g .
i to
f u n c t i o n g~
of the d i s c r i m i n a n t
6 V
corresponding
to
--~,Sb )
{x~S
,b(g~(x')
and
- , b I 6 _, u(x') {xES
take
for Thus
proper
E' A'
= O}
contains = O}.
is thin 9
the two sets
g'
separates
the p o i n t s
of
dense
that
(note that,
S - A ~
~
(~g,bn~v,b) 5, w e m a y
~' - A'
is
it is a f i n i t e c o v e r i n g .
function
l
on
set o f p o i n t s
C
n
s u c h that:
x's
-
A' ,
1
~-i(x') ;
for the g i v e n p o i n t a, -i for a n y c ~ a, c~ ~(a). Suppose
while
gv ~:
(ii)
sheets
=
Since
a linear
(countable)
,
!
and a local hcmeomorphism,
for a
g ~ I ( S b ) _,
B e c a u s e of P r o p o s i t i o n
the z e r o s of
We now choose (i)
such that
~ : S - A ~
l(a)
is d i f f e r e n t
~' - A'
b y the C o r o l l a r y
from
is a c o v e r i n g
to T h e o r e m
l(c)
of
k
3 and P r o p o s i t i o n
8
52
: S ~ ~'
is an open map).
holomorphic symmetric of
functions functions
~. T h e y have
on
Let
~I ..... ~k
~' - A'
which
of the values
holomorphic
of
1
be the
are the e l e m e n t a r y on the
extensions
to
fibres
~'
We
set then
this
latter
k P(~,x')
= ~k +
a
Clearly,
P(l(x) ,~(x))
set being
dense
Since
in
= O
on
by assumption
coordinates
at
neighbourhood is a simple
a
on
of root
on
S) = O
hence,
This
x I - a I .....
is
(a).
- a form P P is a h o m e o m o r p h i s m in a
hence,
of
S - A, S.
S, ~
a;
(x')~ k-v v
v=1
x
b y our a s s u m p t i o n
P(~,a'),
local
(ii),
l(a)
and hence
0~(l(a),~(a) ) ~ O. To c o m p l e t e prove
the p r o o f
(c) ; to o b t a i n
to a p p l y
(c)
We have such that
the second
part
to a n e i g h b o u r h o o d
(e)
holds
- A',
and
with
~(x ( j ) ) - = x'.
is c l e a r l y
5, it suffices of
point
functions
of
~
and (d) h o l d s on S - A. Let (1) (k) x ,...,x be the points of
let
Consider
of the
n o w to
(b), we have
of a given
o n l y to find h o l o m o r p h i c
x'~'
this
of T h e o r e m
only S.
on
~'
S
-
A
the sum
form
k-1
~ (x,)~v V ----O
where ~v(x')
the
~
are h o l o m o r p h i c
is a linear
coefficients
combination
depending
on
~' - A',
of the(
o n l y on the
l~v(X')~,<
M max 3
and 3))
x
lh(x(J))l.
for any with
Hence
Xle
53
In p a r t i c u l a r ,
the
is b o u n d e d
S')
to
~'
on
such
are and
that
(e)
bounded
a holomorphic
holds.
in the
If
substitute
- A'
1 in
points
are
dense
dense
subset
of
S - A;
The
set
irreducible universal Of
S
there
is
course,
hence
this
simpler
we
where
n,o any point
corollary
fashion.
In
on
= ~p
of
fact,
such
regular
that
which
proved
the
on
a
S' is
last
= O.
~o
is
is a
sufficiently be
~-l(x')
(l(x),~(x))
I(So)
can
that
(d)
is n o t
set
f~
points
holds
P' (x)
such
obtain
S
an a n a l y t i c
at
the
(d)
set
exists
denominator
somewhat
extension
identity
is a p o i n t
S - A,
in the
If
x
separates
of p o i n t s
contained
Corollary.
where
and
such
clearly
h
V=O
~ = l(x)
= x'~'
Remark.
(since
k-I
p(~;x') - xlJll'"
j=1
~(x)
~' - A'
so a d m i t s
k
we
on
near
O.
directly part
in a
of o u r
a r g u m e n t s s h o w s t h a t w i t h t h e n o t a t i o n of P r o p o s i t i o n 3, ~P p+l is s u c h a u n i v e r s a l d e n o m i n a t o r . ~x p+l W e s h a l l see l a t e r t h a t the f i n i t e n e s s of t h e f i b r e s of
~ IS
is a c o n s e q u e n c e
of
the
hypothesis
that
~ IS
is
proper. Remark. that
Theorem ~
: X ~
4 remains
~' - A'
is
assumption
that
~
: X ~
fibres
that
X
has
we
and
have
seen
subset and
~
B'r : X -
covering. be
done,
~' - A'
such
Then, first
applications
of
the for the
A'
is
dimension
proof
~
if w e
that
X -
constuction - A'
continuation
of -
at
each
5 that
is the
B',
assumption
by
map
the
with
point.
there
~-I(B')
B'
the
covering
a proper
p
of T h e o r e m
~' - A '
x'~'
replace
an u n r a m i f i e d ~'-
in the
~ - 1 (B')
valid
finite
In
is
is d e n s e
fact,
a thin in
X
an u n r a m i f i e d functions
and,
theorem,
by
Pf(x)
can
successive
extended
first
to
54
x'~'
- A'
then to
x'~'
The rest of the p r o o f
remains
the same.
Proposition
IO.
analytic
set in
Let
be
S --a
constant
(Maximum Principle) ~
and
irreducible 9
on
S
of
a neighbourhood
of
so c h o s e n
U = U'
that
Proposition
and s u p p o s e
f(a)
We m a y s u p p o s e
: SnU ~ U'
a holomorphic
that
O,
of
be an
function f
x U" , U'c
a.
Then
f(a)
= O.
f(S)
Let
C p , p = dim ~ o
finite
x'~U',
~.
is not
let the c o o r d i n a t e s
has
on
C.
a = O,
and
is proper, 3). For
in
S
that
in a n y n e i g h b o u r h o o d
is a n e i g h b o u r h o o d Proof.
f
Let
6(x')
fibres
~ O,
in
U
be
Cn
be
and
(and s a t i s f i e s
let
al,...,a k
be the e l e m e n t a r y s y m m e t r i c f u n c t i o n s of the v a l u e s of f -I on ~ (x') nS ; then the a admit h o l o m o r p h i c e x t e n s i o n s v to U'. Let k P(~;x')
av(x,)
= ~k +
k_v
~=I Then,
for
x'~U',
6(x')
/ O,
f(~-i(x') ) = { ~ C I P ( ~ , x ' ) of the roots f(~-1(x'))
= {~CIP(~,x')
~ O
(since
near
f(O)
variable
= 0}.
of a polynomial,
By assumption, ak(x')
in
we h a v e
O~U'.
for a n y
f~I(~o).
Further,
By t h e p r e p a r a t i o n
that
there
Hence,
by Lemma
5,
(O) = 0 for each v v a f t e r a linear c h a n g e of
a k ( O .... ,O,Xp)
theorem,
x',U'.
a
= O). We m a y suppose, C~
b y the c o n t i n u i t y
we h a v e
= O}
we h a v e
Hence,
~ O
exists
near
x = O.
a distinguished
polynomial m
Q ( X p 9 1 .... ,Xp_f~)
such that
P
and
Q
= x
TMp
have
+
b u (x I, .... Xp_ f~)x m-vp v=l
the same
zeros n e a r
O.
But clearly 9
55
since O
b
(O) = O, for a n y ~ v Q ( X p , O .... ,O,~) = O;
with
for w h i c h
there
a neighbourhood f(S)
exists
x'
of
Since
O.
is a n e i g h b o u r h o o d
Corollary
i.
of a f i n i t e Proof.
number
f
on exist
if
such
v ~ U,
passing sv ~
the
set
O
analytic
~f(S)
SoES,
since
there
are
infinitely
Hence
f
that
a v ~ ~.
S
is c o m p a c t . many
sv
constant
Let
in
Then,
Cm.
S
on t h i s
f(s o)
~
in an o p e n y ~ C m,
a compact
analytic
subset
Corollary
3.
Let
~
of
be
follows
set
10.
Proposition
II.
is i r r e d u c i b l e . U
such
connected.
in
~
and
Let Then
that
n
consists
the
S a
infinite,
suppose
~ s O = U_Sk,so , least
and
one
that then
~ k , s o.
so t h a t
f(S)
then
holomorphic
set in
and set
A'
map
in
~I~
be has
2 and
the
C n,
of an
: ~ ~ Cp
be
is p r o p e r , p
subset of
then
at e a c h p o i n t .
remark
an a n a l y t i c
preceding
set in
a fundamental
set o f r e g u l a r
x
a thin -I ~ - ~ (A')
of d i m e n s i o n
from Corollary
Proposition
hoods
= ~(~)
at a n y p o i n t
is an a n a l y t i c This
~'
an o p e n
is an a n a l y t i c
p
C
cn).
of
dimension
= O},
set in Cn i n t o an o p e n set -I ~ (y) is a f i n i t e set (being
Let
X
in
component,
= a,
the p r o j e c t i o n . If
is
= (~CIP(~,x')
we may
If
be a p r o p e r
for
~'
O
a ~f(S) a / v ' v s v 6 S , f(sv) = ~v; by
on at
a contradiction.
(f(S) 1O,
set
= O
If it w e r e
if n e c e s s a r y ,
a~f(S)
analytic
~C
near
and
Let
of
2.
p near
for a n y h o l o m o r p h i c
(f(S))o
is a n e i g h b o u r h o o d
Corollary
x
P(~,x')
S
set
is f i n i t e . -
to a s u b s e q u e n c e
-
with
is
O.
f(S)
that
is n o t
hence
f(~-i(x'))
to p r o v e
C n,
there would
there
of p o i n t s .
It s u f f i c e s
function
O,
near
of
A compact
near
~
system
points
of
such
that
S --a
of n e i g h b o u r S
in
U
is
56
Proof.
Choose
further,
with
Sn{x~U'!6(x') that
U
such
the n o t a t i o n ~ Oj
if
X
is n o t
of
X,
~I Y
is a l s o
on
Y,
= i
on
of
S,
S' - X
extension f~I(S_o)
such
for a n y
x'~U',
(viz x~Y) F
points which
F(x)
arbitrarily
Corollary.
= O.
a
are
is a n e i g h b o u r h o o d
This
6.
in
and
Cn
of
S
let
there
many
holomorphic the
be
O~S.
exists
set of
set
(b)
of Proof. U' if
f
l
there 5,
5,
on
~o"
is
xs
This
proves
of
such
Now,
there
where
are
f / O,
Proposition
II.
d i m --a S = p,
that
is
so t h a t
X - Y
and
points
there
Fs
on
O
= O
a holomorphic
But c l e a r l y
from
S
Proposition
an a n a l y t i c Suppose
functions singular =
that
fl
"'"
... = f
m
set
has
then
dimension
ii a n d
in an o p e n
set
in
V
such
that:
m
of
SnV
(x)
= O}-
is p r e c i s e l y
denominator
at e v e r y
the
point
V. Choose
U = U' with
p
Proposition
S is i r r e d u c i b l e . --o V o_~f O and finitely f
'
points
is a u n i v e r s a l
the c o o r d i n a t e s
= { x ' , c P i , x ' , < e'},
proper
has
to T h e o r e m
a nei~hbourhood
{x~Vlfl(x)
each
set of r e g u l a r
O.
'
(a)
h
as above, V
at o n c e
Let
Then
be t h e
VnS.
follows
Theorem
functbn
h
By L e m m a
to
Hence
Let
is the
near
covering.
component
clearly
F I(So).
S,
Now
is h o l o m o r p h i c
~ O,
close
that If
at a n y p o i n t
S'
to p r o v e
is a c o n n e c t e d
and hence
at all p o i n t s
implies
there
If
F = fh
6(x')
with
vanishes
Y
and
set
It s u f f i c e s
By the c o r o l l a r y
that
the
is a f i n i t e
a covering.
is thin,
4 is v a l i d
is c o n n e c t e d .
and
X - Y.
S'
SnU.
= O}
connected,
to
in
/ O}
{x'~U' 1 6(x')
Proposition
as b e f o r e ,
is d e n s e
X = Sn{x~U!6(x')
: X ~ U' -
that
• U", finite
then
U"
( x l , . . . , x n)
= {x"~Cn-Pl.x",
the p r o j e c t i o n
fibres.
in
C
< e"}
: U n S -~ U'
n
and
such is
that
5.
57
Let
e
> O
be
i = i .... ,p,
sufficiently
small,
and
set,
for
I sij I < ~,
j = i ..... n, n
i Then,
if
e
linearly
is small,
independent.
enough 9 Let Let be
Ua
the m a p
K'
in
~a(x)
then
and
~(x)
subset Because
subset
of
it c a n n o t
be
in
' X p + l .... 'Xn
< Q'
be
enough; U,
~~
let
To p r o v e U I',
adherent .
and b e i n g
closed
e
we
in
I to P r o p o s i t i o n
is s m a l l
is c l o s e d
if
U,
small
remark
to a n y p o i n t
Further,
e
: S n U ~ ~ U I'
If
this,
(K')
~s
are
U I = U i' x U" .
and
and
and
~
that
in
of
in
if
U,
~U
(x)~K',
neighbourhood of U I' -i ~s (K') is c o n t a i n e d
hence
of C o r o l l a r y
fixed,
,...,~p(~) (x)).
U
in a c o m p a c t
of
Qi
is p r o p e r .
is c l o s e d
lies
is s m a l l
Let
= (~)(x)
~
clearly
U
j
'''''
I I~ i(a) (x) i < QI } ,
is a c o m p a c t
since
~i
~ij
U I' = {x'~U'l Ix' I < Qi }
= {x~U
enough,
j=l
then
U
if
e
in a c o m p a c t
is i t s e l f c o m p a c t . -I Ii, ~ (x') is S
finite
for a n y
Now,
for
x ' ~ U I. any regular
exist
~ij' laijl
system
of c o o r d i n a t e s
O
such
that
small). chosen
If
the
of
set of
of h o l o m o r p h i c denominators of
S
follows replace
in
and (~) ~
S
of
that
a. L e t
S
in
U,
~s)" , . - . , ~ ) W
be
(such a
form
exists
at
coordinates
there a
a neighbourhood
W
is r e g u l a r
form
a, at
if
let a.
of
~ a
There
is be
so
exists,
P' w h i c h is a u n i v e r s a l d e n o m i n a t o r s P'~(a) / O, w h i c h v a n i s h e s on the
Us, S
in
U s.
functions at a n y p o i n t
W
is g i v e n
easily W
at
a
5, a f u n c t i o n
at a n y p o i n t singular
such
l~iJ 1 <~ U s
a~WnS
that
by Theorem
We
< e,
point
from
by
this
by a smaller
Hence,
{ft} of
in W
there W
which
such
that
{x~Vlft(x) and Chapter open
set
exists
are u n i v e r s a l the
= OUt}. II,
V).
a family
singular Theorem
Theorem
5
set 6
(if w e
58
C o r o l l a r y I. Let S be an a n a l y t i c set in an o p e n set n ~2 in C T h e n the set of s i n g u l a r p o i n t s of S is an analytic
set in
proof.
Let
~.
a~S
the i r r e d u c i b l e analytic
dim
S = US --a --~,a'
components
S
are
B~,a
in the u n i o n
b y the C o r o l l a r y small that
for
In p a r t i c u l a r
that bs
to P r o p o s i t i o n
~u
on
S
S
with
S
of a i n d u c i n g
S
o
~v,a
and n o n e of t h e m is con-
if
b)
is c o n t a i n e d
T
N ~
for
u. Hence, W
so
~ / u.
in the u n i o n of
(SvnS ~) .
In fact,
at a n y p o i n t of since
at r e g u l a r then
Sv
b
follows S
so t h a t
from this and T h e o r e m
the d e c o m p o s i t i o n S, S
of
~o
o.
If
f
for a n y
b~S
nU,
on
N / v,
b
and if
g
If
The c o r o l l a r y
6. S
--O
= U S
is
B~,O
components,
of t h e s e g e r m s
is a u n i v e r s a l
If
is s i n g u l a r
S u.
into i r r e d u c i b l e
be r e p r e s e n t a t i v e s
h o o d of
and if
T u.
set is c l e a r l y
is s i n g u l a r .
Sb = S
is n o t i r r e d u c i b l e
--O
it is c l e a r
not on
an a n a l y t i c
point,
S
of the set of s i n g u l a r
U
if and o n l y if it is s i n g u l a r on
If
let
~
then
(S~nSu)
clearly
be an
II, we m a y c h o o s e
is the u n i o n
P
u ~ ~,
irreducible b4
are
We c l a i m t h a t the set o f s i n g u l a r p o i n t s o f
is r e g u l a r ,
S --~,a
--u,b
S
S
Su
(dim -S- p , a' d i m --~, S a)
no g e r m
lie on
p o i n t s of
W
the
of the o t h e r s , w e h a v e
'
which
Let
b ~ S u nS nW w e h a v e < m i n (dim S ,b' d i m S
(S ,bnS b)_ ,
the o t h e r s .
where
_Sa.
irreducible
(Su, nS a ) < min -a --~,
dim
of
set in a n e i g h b o u r h o o d
S i n c e the tained
and
in a n e i g h b o u r -
denominator
is h o l o m o r p h i c
for
on
S
u,b g = O
U,
U
S ,
is a u n i v e r s a l It f o l l o w s Corollary hypothesis
g / 0
on a d e n s e
denominator
s u b s e t of
at a n y p o i n t o f
S u, SnU
then for
f r o m this that w e h a v e 2. that
Theorem ~o
6 remains
valid
is i r r e d u c i b l e .
if w e d r o p the
S.
h = gf
59
Corollary Q and
in
C n,
if
there for
3.
If and
aES, is an
S
Hilbert
is
the
then
for
any
k
) I
integer points
follows
an a n a l y t i c
subset
holomorphic
function
function
F
Proof.
Let
on
singular
points
set
of
S,
w h i c h v a n i s h e s on --a' A ,a such~ t h a t fk is a u n i v e r s a l a. from
The
only
Corollary
change
isolated
point
2 above
and
the
F
We m a y
the
if
C
n
and
2.
there
is a u n i q u e
Then,
it e x i s t s , for
for
Ac~ any
holomorphic
that n C ,
in
is o b v i o u s .
every
a~A,
holomorphic
suppose
in
V
a = O, that
there with and,
O
is
after
is an
set
i
:
9
sufficiently
that,
in
( n -
if
that
an
a~ is c l e a r
F,
of c o o r d i n a t e s of
set
F I ~ - A = f. of
and
An{xlx
is
an o p e n
~ - A,
to p r o v e
V
a linear
be
on
uniqueness
we have
> O
~
with
F I V - A = f l V - A.
it
in an o p e n
f~n
of d i m e n s i o n
~
a neighbourhood
Q
of
set
Nullstellensatz. 12.
If
set
near
at o n c e
Proposition
Hence
is an a n a l y t i c
A
at all
This
S
:x
--0).
n-2
small,
and
( O , . . . , O , ~i, 0),
=
~
..
is
large,
the
set I
-Du =
~x~C n Ix I = . . . = x
n-2
Ko =
{x~cnlxl
= Xn_ 2 = O,
= O,
x
n-I
=--,
IXnl
~ Q}c~-
A,
and
Hence,
by Chapter
neighbourhood in
V
such
connected, continuation
V that
it
= I, of
O
F = f
follows that
Proposition
12,
(containing near
from
F = f
Xn_ 1 = O,
the in
K o.
There Ko)
V - A.
is and
Since
principle
IXn l= Q } c n
a connected F
V - A of
- A.
holomorphic is
analytic
60
Proposition in
Cn ,
for
and
a~S,
S --a
13. let
f ~ O
induced
we
Let
by
S
be
an a n a l y t i c
f
be h o l o m o r p h i c
on
any
S
at
in
irreducible
a.
Then,
in an o p e n ~.
Suppose
component
if
S'
=
set
of
that,
the
{xs
germ
= O},
have d i m --a S' = d i m --a S -
Proof. by
i. W e
the
first
preparation
suppose
theorem,
that
I.
S = ~
we m a y
and
suppose
a = O.
Then,
that
P f(x)
= xP+ n
a~(x')xP-U,n
x'
=
(xl, .... Xn_1) , a
(O)
= O.
v=i Clearly,
if
hood
0
of
x n ~U" is g
U = U' such
then
=- O,
{since
definition
projection
hence g
if
~
clearly
: S'
and
that
--oS
dim
that
~
Further,
is d e n s e
6(x') the U
in
~ O,
is
if
SnU.
covering,
points is
U = U'
: S n U ~ U'
fibres 9
finite
general
S'
of
lying
S
enough)
also
denoted
h(x')
= O~
by
h,
is the
is as ~
say,
be
imply
{x~Ulf(x)
that
= O} ~ U'
the
so that,
suppose
p = dim So,
of
O,
U'r
of
x'
-
so a d m i t s
to
U'
that
we U"c
can Cn-p
and has
2,
Let, the
for
WnS'
find such
finite
values h
~ O} = O}
of
f
at
is b o u n d e d
{x'~U', onto
is
x'r
a holomorphic
Clearly
a = O
extension,
6(x') U'.
(if
/ O,
Further,
J
if
x~S,
xv~S, hence
6(x')
= O,
6(x' v) ~ O, h(x'v) ~
0
we
xv ~
can x.
so t h a t
find If
by
W = {xs
{x'~U'16(x')
Then
and
of
Cp
proper,
in L e m m a
product
projection
then
i.
If
sheets
over
~(S')},
we may
: W ~ U' q
on
= n -
surjective,
Now
h(x')
small
x U"
6
of
is a n e i g h b o u r -
O
case,
is i r r e d u c i b l e 9
a neighbourhood
=
vanishes D
the
, U" c C,
g = g ( x I ..... Xn_l) ~I(So) ,
of d i m e n s i o n , 2. F o r
n-i
x ' ~ U ' , f(x' ,x n ) = 0
that
the
surjective;
• U" , U' c C
a sequence
f(x)
h(x')
= O,
= O.
of p o i n t s
then
Hence
f(x
) ~ O,
a
61
x(S') r
= 0},
the c o n v e r s e Since
inclusion
f ~ 0
on
S
in
of
Cp
By the
after
(and s h r i n k i n g
Xp_ i
C p-i
of
U')
Then
9
since a n y f u n c t i o n
vanish
on
T,
dimS
=p-
Theorem The
that
7.
Let
o-
S
neighbourhood
~
in I.
following
holds:
of the p r o j e c t i o n = x p-1 (U')
"
must
so that
application 9
Sv
set in an ope n set in on
of h o l o m o r p h i c
and
of c o o r d i n a t e s
g (xl, .... Xp_l) ~I (_So)
functions
~ f, u n i f o r m l y on c o m p a c t p h o l o m o r p h i c on S. aeS
U'
analytic
f
Let
to
important
be am
is a s e q u e n c e
Proof 9
given
= 0}.
•
another
space of h o l o m o r p h i c
(fp}
that the
I(S_~) n~D_ 1 = {0},
i =dimS 5 has
argument
T = x p-I (x(S'))
It f o l l o w s
Theorem
same way,
= (x'EU'h(x')
a linear c h a n g e
be the r e s t r i c t i o n
onto
x(S')
in the
9
we can suppose,
Cp
let
, so that h ~ O.
~0
above,
and one proves,
S
is complete;
functions
subsets
of
on
S,
S
C n.
i.e.
if
and
then
f
is
be a n a l y t i c
sets in an o p e n k --aS = 9=kJl S a is the
of
a
such that
S
into
irreducible
S
after
--9 t
decomposition Theorem
of
--a
5, a p p l i e d
ordinates
to
Cn ,
in
--9,a
there
and a h o l o m o r p h i c
is
components.
a linear c h a n g e
a neighbourhood
function
u
in
V
9
is h o l o m o r p h i c ~
in
we h a v e
V9
on
V nS, U
there
such that 9 w i t h
{x~V nS 9
9
u
(x) fi O}
M'
V9
of co-
of
such that
a
in
if
V
is a h o l o m o r p h i c
function
> 0
of
independent
is d e n s e
in
9
Uv~ = ~
By
in
~,
V nS V
9'
VgnS9,
and ti~vilvv ~ M' ii~11VvnS. If
gv
is
a holomorphic
function
in
V
9
vanishing
on
~/g
S~ ,
62
while
{x~Selge(x) ~ O}
exists,
by Proposition
for any ~e
~
9'
if
holomorphic on
(= g e ~ )
here
is dense in
on
V
V
is
e
V uS,
SenV e small
(b)
ll~ellve ~ MII~IIVenS;
Let
and
M > O
~ O}
~
on
V e uS,
is independent of
is dense in
~a =~R,a
functions on
then,
with v e ~ = ~e
{x~Senvelve(x)
enough),
ge
there is a holomorphic
e
(a)
v e = geue
(such a
@.
Further,
SenV e-
be the ring of germs of holomorpHic
at
a,
and
of functions ~vanishing on
~a = 5a (S),
the ideal in
S a.
Consider the homomorphism ~ : ~a~% given by
~(f) = (vlf '' .. 'vkf) "
By Cartan's theorem if a sequence
h
say) and
Let
Ea = a(%)
+ ~ ka r
a"
(Chapter II, Corollary 1 to Theorem 5),
(hp)
in a neighbourhood (to
k
of U
k-tuples of holomorphic functions of
a
(hp)a~Ea
converges uniformly on for each
p,
then
U
(h) a~Ea.
Let
{f } be a sequence of holomorphic functions on P which converges uniformly on ~nS. Let ~o = fo' ~p = f
p
- f
p ~ I.
p-l'
We may suppose that
co
ll~pllQnS < co.
By our remarks above, there are holomorphic
p=o functions
~p
(e = 1 ..... k)
on
V
,e on
then
V enS,
V
ll~p,eIIve ~ MII~IlvenS"
(a(~p))a-
converges on
(~p) a ~ k,
V = NVe
since
If
hence
such that ~p =
(~p,l ..... ~ p , k )'
(0p)a~E a.
ll~pllv ~ M
Ve~p = ~p, e
Further,
ll~pll~nS ~ co.
63
Let
~ = (~(I)
(@)a~Ea,
i
o
9
9
,~(k) ) =
so that
a holomorphic
there
function
@p
Then, by
"
our r e m a r k
is a n e i g h b o u r h o o d ~
on
W
W
such that
of vv~
above
a
I
and
(v)
=
on WnS. Let we have for in
vvf = ~ ) o n
~ = I .... ,k. V nS,
so that this
f = lim fp =
this ~ = f
proves
Remark.
~p. WnS,
on
so that
But since
implies WnS.
Then,
that
since
Vv~p
vv(f - ~) = O
{x~VvnSvlvv(x) f = ~
Since
@
= @p,v'
on
# O}
WnS~
on
is dense
for each
is h o l o m o r p h i c
on
the theorem.
The
Bungart-Rossi
idea of this p r o o f 7.
is e s s e n t i a l l y
WnS
that of
~, W,
64
CHAPTER
IV. - C O H E R E N C E
In the c h a p t e r s theory
of sheaves
sheaves
here.
in C h a p t e r sheaves
All
(Theorem
S
We have
defined,
Let a subsheaf
of the
is c o n t a i n e d
complex
theory
analytic
in Oka's
of
sets
theorem
there
to
fb
if there
exists
S.
set
sheaf
and
X on
(Y,%)
to
~X,b
X).
if to e v e r y
(U, ~IU)
is
of c o n t i n u o u s =
(OIU)b
if and o n l y
fb = g~(b) ~ sections in
U.
~
of germs
A continuous
are c o m p l e x
on
a homeomorphism
that
sheaf
~ =~X
functions
that
functions
(or the
S.
and an a n a l y t i c
exists
a germ
on
and
space
the c o n t i n u o u s
holomorphic of
such
such
UcX,
space,
U
way,
is a sub-
functions
a complex
beU,
belongs
set
~S
of c o n t i n u o u s
there
g~(b)~S,e(b)
functions
(X,~ X)
i.e.
The
in an o b v i o u s
topological
Cn
n C .
in
of germs
(deS).
of c o n t i n u o u s
in
for
b)
are c a l l e d
structure
~
~
~S,a
Further,
is a n e i g h b o u r h o o d
(at
any open
S,
carries,
is c a l l e d
such that,
holomorphic
on
on
of germs
(S,~s),
functions
For
a
of germs
in an open
: U ~ S
at
sheaf
set
III , the ring
U ~ deS S,a
(X,O)
aeX,
isomorphic
where
this
13.
set in an open
be a h a u s d o r f f
The pair
the
=
S
sheaf
X
S
functions
of a sheaf
of the
U
with
the
of c o h e r e n t
The general
except
in C h a p t e r
O = O
structure
over
34.
exclusively
be an analytic
collection
set
is n e e d e d
in C o d e m e n t
functions,
of h o l o m o r p h i c
point
also
properties
We do not d e v e l o p
that
paper
freely
spaces
Let
X.
shall use
in particular,
the m a t e r i a l
deal
we
3).
Complex
sheaf
follow,
of rings).
is d e v e l o p e d
and h o l o m o r p h i c
the
and,
I of Serre's
We shall
•
that
(over a sheaf
theory
THEOREMS
map
spaces,
of is c a l l e d
of ~ : (X,~ x)
is c a l l e d
(y,Oy),
65
holom0rphic
if for any
as
and any
g~y
we have
,~(a) '
g~ induces this
a homomorphism
notion
of h o l o m o r p h i c
of h o l o m o r p h i c and
Y
ring. with
m a p given
are analytic Note
isomorphism
C n,
Cm
Then m o s t
Xa
= ~(X --u ,a ),
then
on
an
if there
such that
X.
are a n a l y t i c
Y~(a)
decomposition
dim --a X = dim --e(a) Y
3.
a~X
is a r e g u l a r
i s a regular
If now therefore
local
(X,~x)
sets
in open
~ : X ~ Y
sets
is b i h o l o m o r p h i c .
introduced
In particular,
2.
~(a) EY
a
~
that we have
~.
is c a l l e d
we have
in C h a p t e r the
III
following.
is the d e c o m p o s i t i o n of the g e r m X = U --u,a X --a at as into i r r e d u c i b l e components, and
corresponding
at
(Y,~y)
is b i h o l o m o r p h i c )
and that
of the n o t i o n s under
~
~
~o~ = i d e n t i t y X, Y
is an a n a l y t i c
can be i d e n t i f i e d
~ : (Y,~y)
respectively,
X
Y --u, ~(a)
a
Y,
X
C.
: (X,~x)
map
the d e f i n i t i o n
If
of
at
~
now that
invariant I.
on
~X, a
functions into
with
Clearly
III in the case w h e n
Further,
say that
a holomorphic
Suppose
are
map
(we also
~o~ = i d e n t i t y
in
sets.
~)X, a"
coincides
in C h a p t e r
mappings
A holomorphic
: ~Y,~(a)
map
that h o l o m o r p h i c
holomorphic
exists
~*
(X,~ x)
define
into
isomorphism
point
if and o n l y
of
of
and regular of
space
X
(X,~ X)
complex
at
Y
~a
the d i m e n s i o n
points
of
X
(S,~s),
C n.
when
space,
of the germ
with
set in
(X,~x)
of
if
Y.
components,
set in an open
for a c o m p l e x
Y (a) --~
is an a r b i t r a r y
irreducible
is the
of the germ
the d e c o m p o s i t i o n
(= dim Xa ) ,
analytic
point
= U --u, Y ~(a)
we m a y of
of
X X
b y using
a
S
an
We w r i t e
no c o n f u s i o n
being simply
X
is likely.
~(a) .
66
We m a y
further
and s i m i l a r X
carries
define
analytic
notions.
Note
a natural
structure
spaces.
proposition
I.
the c o m p l e x
space
a~X,
a
Proposition space
X
dense
in
~ : X ~ Y
X
complex
~*
The
set
and,
3.
If
space
component
for a n y
X
space
X.
@
Further,
is not c o n s t a n t
X (a~X), --a
then
if
of a c o m p l e x X - S
dim S a
on some
is
< dim ~a.
function
f(X)
if
~(a).
points
we h a v e
Then,
-I is f i n i t e
of
of s i n g u l a r
m a p of
Y.
a
is a h o l o m o r p h i c
which of
X --v,a
f
at
some of them.
point
a~S,
of
III e x t e n d
be a h o l o m o r p h i c
set in
A
space.
: ~ Y , ~ (a) ~ X ,
S
space,
subset
in C h a p t e r
into the c o m p l e x
is an a n a l y t i c
Proposition
proved
is an i s o l a t e d
2.
X
of c o m p l e x
We state
the h o m o m o r p h i s m
and o n l y if
of
Let
of a c o m p l e x
that any a n a l y t i c
M a n y of the t h e o r e m s o n c e to c o m p l e x
subsets
on the
irreducible
is a n e i g h b o u r h o o d
f(a) . Further,
as follows
limit of a s e q u e n c e uniformly
from C h a p t e r
of h o l o m o r p h i c
on c o m p a c t
subsets
III,
Theorem
functions
of a c o m p l e x
which
space,
7, the converges
is a g a i n
holomorphic. Corollary following
5 to T h e o r e m
hood
U
theorem.
a holomorphic such that,
for
dim
x
Let
map. x~U,
f-lf(x)
space
into i r r e d u c i b l e
be c o m p l e x
X, Y
Then,
any
a~X
spaces
and
has a n e i g h b o u r -
we h a v e { dim
We shall n o w turn to the g l o b a l complex
III g i v e s us the
important
Semi - c o n t i n u i t y f : X ~ Y
3 of C h a p t e r
a
f-lf(a). decomposition
components.
of a
An a n a l o g o u s
theory
67
does
not
exist
sets
in an o p e n
more
closely
in the n e x t
Let
be
X
is c a l l e d
A
in
X,
for
1.
C ~ A
be
i__nn X
an i r r e d u c i b l e
locally
irreducible for
some
analytic
an a n a l y t i c
subset.
of
A = BuC; A
X
subset
of
X = U X a.
subset
of
X
such
let X'
closure
analytic
B, C
is i r r e d u c i b l e .
T h e n a if
the
sets
if no
space r and
X.
and
analytic
say that
X'
finite,
problem
exist
that we
points of
analytic
l o o k at t h i s
A
a complex
component
is
space,
such
X
(or r e a l
chapter.
is p o s s i b l e ,
Let
spaces"
shall
if t h e r e
connected
{Xa}
We
a complex
set of r e g u l a r
is
analytic
Rn).
reducible
decomposition
the
set
B ~ A,
Theorem
"real
X'
be
is a n y
= X' X.
of The
Further,
X' family
any
is c o n t a i n e d
in
Xa
a. Let
Proof.
isomorphic
a~X
and
let 9)
(by a m a p
V
be
a neighbourhood
to an a n a l y t i c
set
of
S
a
in an o p e n
O~S b e the p o i n t c o r r e s p o n d i n g to a, Cn . Let k be the d e c o m p o s i t i o n of S into a n d let S = J S --O --o v =I --v, o irreducible germs. Let U v be a n e i g h b o u r h o o d of O such set
~
in
that
the
set
connected Since
S'
of r e g u l a r
V
and dense
S'n v
connected
~. S v~ ~ ~ complex
in
points
S v n U u.
S'
S nU v
.
Let
W
V
subset S"
V'
dense
S
in
U
--
~~ ,v
S" = S vI
Let
is an a n a l y t i c manifold
of
/ S' v
is a g a i n
S
9
of the connected
and
V
be the n e i g h b o u r h o o d
of
a,
v
for w h i c h
is
V
We V
-1
~(W)
= ~
U v = U.
Consider
the
set
~
II
(S~)
= X~.
U I, Xv
Clearly connected X"nX' over
is c o n n e c t e d X a'
component
= @.
Hence
those
v
X
with
and of
X"r X' ,
nW = X ' n W X"nX'
X'.
V
Hence,
we have
X "vr
is the u n i o n
/ ~.
Hence
~(X
given X'
of nW)
a
or X"nW is the
V
union
of
S"nU
over
those
u
p
S"nU
= S nU,
V
U
with
X"nX'
~ ~.
But
since
V
this
latter
set
is c l e a r l y
analytic
in
U;
68
hence for
X
is an a n a l y t i c
5
at m o s t
components family.
k ~'s of
Suppose sets B
A'
B'
are
X'5
analytic
so
A'uB',
we have
X, then be
those
finite
is
Y =
~J
and
This means
that
Definition
I.
irreducible Corollarv
subset Proof. if
A'
X'
I. X' 2. X,
= XtnA,
- A'r
= X
nA.
5.
Let
a~X
.
Y ~ A,
in e i t h e r where
But,
since
subset
then
A
A',
B'
X' of and
analytic
If
A~X
a~A, Then,
and
for let
since
is a n a l y t i c
subset
of
any
~,
~I .... ,5
{Xa} and
p
is l o c a l l y
Y ~ A,
since
in
A = A1u...UApUY.
we h a v e
A = Ai
for
Since
A i.
some
Ar X 5 i .
The
X~
defined of
A complex
Since
and d e n s e
analytic
a contradiction.
we have
components
set
of
are
any analytic
X'-~, b u t
X~nA
A i = AnX~i,
irreducible,
Corollary
B
X' ~ X'.
irreducible
5
finite
~i i=I ...,p
If
if the
A
for w h i c h
' as
is an
for e a c h
5
in
irreducible
X ' = A'uB'
Hence
dense,
A = U A A
A
A
A,
contained
manifold,
dense
X nW /
irreducible.
is n o t
of
of
a locally
where
subsets
complex
if
is
Xs).
are n o w h e r e
Finally,
form
X5
X'
that
is the n u m b e r
= AuB,
in
is n o w h e r e
It is c l e a r
{Xs}
that
X
dense
is a c o n n e c t e d X57 ~'
now
Clearly,
(being
k
Hence
that
~ X a.
or
where
X . --a
We p r o v e
set.
X
space
X'
X - A r X'
is i r r e d u c i b l e
points
then
is d e n s e
A' / X', X',
X
of
X
is i r r e d u c i b l e
A ~ X,
X - A
the
in
= X,
also
and
X,
A
we have X'
in
X - A
- A' X.
if a n d o n l y
is c o n n e c t e d . is an a n a l y t i c
is c o n n e c t e d
so t h a t
and hence - A'
are c a l l e d
X.
of regular If
above
X'~
and d e n s e A.
in X.
Hence
is c o n n e c t e d Since
is c o n n e c t e d
and dense.
69
Corollary of
3.
If
X
irreducible,
dim --a X
is i n d e p e n d e n t
a~X.
Proof.
d i m --a X = dim --a X'
is a c o n n e c t e d from C h a p t e r Corollary
III,
4.
Proposition
If
X
s u b s e t of
Proof.
For any
(by C o r o l l a r y ~ ~
and
contains
2). Z
then
a~X,
A
Since
if
subset
contain
any i r r e d u c i b l e
follows
from C h a p t e r
and
Y
5.
If
then
Y
irreducible Proof.
Z --a
of
subset of
if the d i m e n s i o n
that
Corollary
If
is irreducible, an i r r e d u c i b l e
is
X
if
4
p
p,
at e v e r y
f a m i l y of
of
X
~ p,
SU
p.
into
Let
Y~c X~.
and C o r o l l a r y
If
X X
has
4 above
5 follows. set in
in
U
Q,
Ur Q,
set
is a n a l y t i c
in
Y
at e v e r y point
is
with
U
Su
of
p
Y~
in
irreducible.
cannot
space of d i m e n s i o n
of each
set
sets
Z
7.
Q,
with
of
and there
germ
a~S U
aES
a~U,
whose
s y s t e m of n e i g h b o u r h o o d s
analytic
A
Corollary
has d i m e n s i o n
for any o p e n
S
then
that
_X a .
of
~a component
~a'
of d i m e n s i o n
Corollary
analytic
is a f u n d a m e n t a l
S --a
it follows
is an a n a l y t i c
S
then
Conversely,
then
dim X
Y~ = X~.
6.
irreducible
Y
component
then
is an
the g e r m
be the d e c o m p o s i t i o n
if and o n l y
p,
at once
X.
components;
implies
A ~ X
set i n d u c i n g
Proposition
irreducible
dimension
and
is the u n i o n of a c e r t a i n
be an i r r e d u c i b l e
X'
< d i m X.
is a c o m p l e x
Y = U Y~
since
is an i r r e d u c i b l e
X,
of
a~X', 3 follows
contain
component
components
Let
dim A
cannot
III,
X
an a n a l y t i c
point,
Corollary
is i r r e d u c i b l e X,
of
6.
is an a n a l y t i c
an open
Corollary
is i n d e p e n d e n t
complex manifold.
analytic
of
is
at
a
and t h e r e a
and
(Su) a = ~a'
S --a is is
~Sa
.
70
Proof.
The first part follows
Proposition component
II and Theorem
of
S'~U
I; in fact,
III,
there is a connected
(S' = set of regular points of
dense in a neighbourhood SU
from Chapter
an irreducible
of
a
component
on
of
S
S)
and we may take for
SnU. The second part
is
trivial.
Theorem
2.
of analytic
Let
X
be a complex space an d
subsets of
X
which is decreasinqly
for the order defined by inclusion. subset
K
Proof. that
of
X,
being evidently
is an analytic
C n.
Further,
a~,
and let
i.e.
I~ = I(A~,a)._
{I~}
contains
I~
A~onU = A~nU.
be the ideal Since
contain
a.
any
A~nA
U
of
I~o,
= A~o
)
,
i e
subsets of
A~ "
The intersection X
is analytic
in
of elements of
we have
the proof.
Now,
I~ )I~o,
and
of
~,a
which
A~o
(~)
A~)~A~o
= ~o,a"
Now,
-
(since
~o
2 to Theorem
for
I,
"
of any family
~
of analytic
X.
Apply the above corollary
intersections
~ > ~o'
an open subset of
-(~) )~0
A~,
the family
so that
components
so that by Corollary
Let
We claim that there
so that
to show that
contains
9
Corollary.
I~o.
I~ = I~o,
in
deffned by
such that if
A~g A~o ,
It suffices
(~)
(a)
Proof.
a
~
S = ~.
is noetherian,
element
of
~ > 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,
Then,
SnU
the
S --o let
and
for
then
U"
~.3 (S)
that
of c o o r d i n a t e s of
0
U"r
C m-p,
in
is a n a l y t i c
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{xcU6(x')
if
= U 1'' • ...
is
from
of C h a p t e r
j ~ 1,
S
III,
is irreducible.
such that
o
We m a y
set in an open
U = U'
= s
to an
satisfies
following
of the p r o j e c t i o n
• U~3 = U.3
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
without real
now prove,
of an open change
which
we
to the case
functions
shall
use
set
Of course, n to that of C
Theorem
3.
(K. Oka)
Let
= O
denote
the
holomorphic
functions.
III,
= p.
of an open
of the sheaf
later.
reduced
~
(S)
sheaf
in an open
be t r i v i a l l y
let
dim ~ ~j
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
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,oX is an
the q u o t i e n t
~n,o
n
~n_l,oXn.
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,oXn,
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
of
131ur
uri~%_l,b,Xn
-
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
relation
at
b = (b',bn),
P 9 = Pl
a (i) (x')x v v n" P=O
(Note:
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+
~ i ) (x,)cl(i) (x,) = O
in ~ n _ i , b''
for each
v;
=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}
is dense in
~ b r ~b"
Then clearly
where
(~p,x'
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,
a distinguished polynomial of degree a unit
SnU,
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)
~
r = Ib (PI ..... Pn-p ) "
Hence,
if
6N~ where
-
N
is large,
A' (Xp+l,-x')
A'
~Ib(P I .... 'Pn-p;6Xp+2 - Q2 ' "'" '6x n
is a polynomial
Applying the
(algebraic)
~%,b,Xp+l~
-
Qn-p ) '
(of bounded degree).
division algorithm relative to
A I,
we conclude that (3.2) 0N~ - A (Xp+l; x') ~Ib(P I .... 'Pn-p; 6Xp+2 - Q2 .... '6x n - Qn_p} = ~ b where
A (Xp+ I,9x') ~ p , b , Xp+ i
Now, for fixed roots; hence
x' , A1
with
has
has degree
6 (x') ~ O,
11
< 11 = deg Al(Xp+l; x') .
PI
has distinct
distinct roots. Further,
if
Ix' - b'l is small, all these roots lie in a prescribed neighbourhood of
Xp+ i = bp+i,
is distinguished.
Hence,
in with
(3.2)
has
11
6(x') ~ O;
for any such
x';
if
since
A i = Al(Xp+ i - bp+l;x')
9' ~(S b) = ~b ~
distinct roots at any hence,
since
again since
deg A < iI,
x'
then
A(Xp+i,x')
near
b'
A(Xp+i;x')
{x'~U' 16(x') ~ O}
is
- 0
80
dense
in
~%,
U'
and
a trivial (3.1)
this
implies
5b c ~ b "
Thus
corollary
implies
that
of
that
(3.1)
(3.1)
5
A ~ O,
so that,
is proved.
and T h e o r e m
is a q u o t i e n t
by
Theorem
(3.2), 5 is
4. In fact,
of the
sheaf
of
realtions
(6N'pI .... 'P n - p ' 6 X p + 2 and so is of finite
Theorem
6.
Proof.
sheaf
The
X = S
type.
Le__~t (X,~ X)
is a c o h e r e n t
- Q2' "'" ,6x n - Qn-p )
be a c o m p l e x
space.
Then
~X
of rinqs.
theorem
being
is an analytic
local,
we m a y
set in an open
suppose
set
~
that n C
in
Then
m
~S
= ~/~(S)
Remark.
IS'
and
so is c o h e r e n t
It is easy
of singular
points
to deduce
from T h e o r e m
of an a n a l y t i c
set
5 that
is again
the
5. set
analytic.
it suffices
a point
a~S
for all
b~U, then b~UnS is s i n g u l a r on S if and o n l y 0f. < n-p. Proof. If b is regular, then i~--~lb
at w h i c h
~a
in a n e i g h b o u r h o o d
is irreducible.
the gj are linear c o m b i n a t i o n s of the fi' we have 0f. 0fi b ~ n-p. C o n v e r s e l y , if rank b ~ n-p,
n - p
at
n - p b;
then
a submanifold of
b;
S = V
trivially Note b
of
of
since near
S.
of the
C
the n
~bC ~b b
that
set
fi'
of d i m e n s i o n and
singular
that we h a v e
rank
\0xj'b
fl .... 'fn-p
= n-p;
of rank
V = {fl = "'" = fn-p = O} p
follows.
is
in a n e i g h b o u r h o o d
dim ~ b = dim ~ b = p'
and the r e s u l t the
say
rank
dim S b = p
since
find
such that
of
exist
we can
)
If
U
there
rank
gl .... ' g n - r
this
S
3 and
In fact,
if rank
to prove
by Theorems
we h a v e
It follows
set of S is analytic in U. 0f. (._~l) ~ n-p at e v e r y point % ~Xj b
81
The above description
of the singular
valid if we suppose that b~SnU,
dim ~b = p
instead of assuming
4. The direct Let
a holomorphic
(Y,~y)
map. Let
i.e. a sheaf of
~I/
of ~
~
3,
is the group f-1(U) .
~
for
is irreducible.
spaces and
X,
f.(~),
as follows,
f.(~)
~ ~},
for an open set
where
i'
is the sheaf
of continuous
It is clear that
= f. (~)
f : X ~ Y
sheaf on
We define a sheaf
over the ring of holomorphic the sheaf
S --a
be an analytic
to the pre-sheaf
over
remains
is constant
be complex
~gX-mOdules.
called .direct image of
UcY,
that
S
image
(X,~x) ,
associated
set of
~U
sections
is a module
functions on
U,
so that
associated with the presheaf
is a sheaf of ~ y - m o d u l e s .
Further,
{~U}
it is clear that the
natural map
F (U,~)
denoting
Theorem 7.
sections of
Let
X
an___~d Y
a proper holomorphic any coherent sheaf image
f.(~) Before
~
over
be complex
map with discrete ~
of ~ X - m O d u l e s
is a coherent
sheaf of
starting on the proof,
is an isomorphism.
U,
spaces, fibres.
on
X,
f : X ~ Y Then,
for
the direct
~y-modules.
we make a few preliminary
reductions.
(a) Y
The theorem being
is an analytic
set in an open set
an open set in
Y
replace
f-l(U).)
X
by
is a coherent a coherent
local on
isomorphic
~
we may suppose that
~
in
Cn
to such an analytic
Further,
it is clear that
sheaf of ~ y - m o d u l e s
sheaf of
Y,
-modules,
if and only if
where
g
IIf
U
is
set, f.(~) g.(~)
is
is the composite
82
of
f
and the i n j e c t i o n
that
Y = 9 (b)
is an open
If
analyt i c
of
Y
in
set in
C
~i ~ ~2 ~ ~3 ~ O
sheaves on
X,
f.(~l )
~.
Thus we m a y suppose
n
is an exact
then the induced
sequence
of
sequence
f.(~2 ) f*(~) ) f , (..~3 ) ~ O
is exact. We have o n l y to v e r i f y
Proof. If
y~Y,
let
~(f, (~3) y.
of
y
and a section
Since
f
is proper,
U
of s a t u r a t e d f-i(y)
= s
on
f-l(W) c L,Vi;
(c) Xo~X'
then
f-l(yo)
by
and O
If
~.
= U U i,
system
suppose U.I
that if
a neighbour-
the
be a n e i g h b o u r h o o d
s'. 1
define f,(~)S'
= S
sufficient
then there
Yo"
a section on
of
y
w it h
s'~F(f-l(w)
to prove that if
is a n e i g h b o u r h o o d
then
g,(~)
Hence,
we m a y
is a c l o s ed
the sheaf on
is c o h e r e n t suppose
X,
then
subspace X'
~'
U
of
in a
that
Then
f: ( ~ )
to prove T h e o r e m
7 for
f'
X
space
X',
by any larger
proper h o l o m o r p h i c
m a p of
of the c o m p l e x
space
obtained by extending is c o h e r e n t
be a p r o p e r h o l o m o r p h i c
f' IX = f.
: X' -+ Y.
X' -+ Y
f
on
ma p w i t h
~---f, ~ ) .
if
%)
W.
is a single point.
outside
f' : X' -+ Y and
of
X
~'
f-l(U)
W
g = f}U,
= {Xo}
(d) X' ,
Let
Yo = f(Xo)'
neighbourhood
inducing
We m a y th er ef or e
then
for which
if
is a n e i g h b o u r h o o d
s ~ F ()f --l (-U )", ~ 3
It is c l e a r l y
such that,
there
is surjective.
U.nU. = ~ if i ~ j. Let V. be a i 3 z s~x a section ~F(Vi,~)__ with xl" and
V i.
--~ F{W,f,(~2 ) )
f,(q)
e v e r y fibre has a f un da me nt al
= {x l,...,xr},
x. and z n e i g h b o u r h o o d of q(s)
Then,
neighbourhoods.
hood of
that
Hence,
X'.
Let
finite
fibres
it suffices
Thus we m a y r e p l a c e can be e x t e n d e d
having
finite
to a
fibres.
xo
83
(e)
f : X ~ Y = tic C n
then there
is a n e i g h b o u r h o o d
an a n a l y t i c hood
U
set in an o p e n
of
Yo
Clearly,
of
onto
a unique ~.
If
U'
x U
in
U,
that
X
U
Cn U,
(f)
~'
above
If
sequence
and
~ Xp - ~ q
that
P r o o f of T h e o r e m U'
• U
7.
proper
and has
-~ O
and there to
is
~
under
of the p r o j e c t i o n Hence, U'
of
we m a y x U,
U'
is the r e s t r i c t i o n
open
to
chosen,
on
~';
there
the
is e x a c t b y
is an
sequence
(b) .
Hence
= %" X
be an a n a l y t i c
fibres;
to p r o v e
subset
~ : X-~ U
of is
the c o h e r e n c e
of
~.(~) at Yo~U, ( ~ c o h e r e n t on X), we m a y s u p p o s e -I (yo) =Xo~ is a s i n g l e p o i n t and that (Xo,Y o) = 0 ~+n. in
We d e n o t e Cn
by
mapping have
coordinates
Yl ..... Yn"
~*
: ~U,O
in
Cm
By C h a p t e r
-~X,O
seen in C h a p t e r
X
={Xo~.
that the p r o j e c t i o n
finite
is
is an i s o m o r p h i s m
s u b s e t of
f - i ( y O)
Let
and s u p p o s e
A
f
-~ ~ - ~ 0
~
f : ~' ~ U
.isomorphic
to
to
and a n e i g h b o u r -
and
x U,
is s u i t a b l y
f . ( ~ X )p ~ f.(~X )q -~ f.(~) suppose
A
and that
on
we m a y
on
=x~,_
isomorphic m
= (x,f(x))
is an a n a l y t i c
in
C
~. (~) = f. (~) .
of p r o j e c t i o n
exact
~
then
in
Ac U'
is the r e s t r i c t i o n
onto
Cm,
sheaf
xo
f (~') r U
set
f - l ( y O)
of
U'
~(x)
an a n a l y t i c
coherent ~
suppose
the m a p
~'
set
such that
proper. fl'
satisfies
by III,
is finite;
III,
that there
x I ..... Xm; Theorem
this
that in
those
3, the
implies,
as we
exist distinguished
polynomials p -I Pv(Xv;
y)
= X u +
ag
v
N=o
(v) which
vanish
vanish suppose
at
on the g e r m of
y = O,
that no
and P
has
X
at
u = i ..... m. a multiple
v a neighbourhood N = D' Dc C n b e i n g connected,
x D D'
of
0
O;
here
the
Further 9
a
(Y) ~",Ou
we m a y
factor. We can find m+n m in C , D'c C ,
= D I • ... x Dm,
where
the
D v
84
are p l a n e
domains
such
that
the
following
conditions
are
fulfilled. (a) a (~)
and
The
are
Pv(xv;y)
induced
by holomorphic
(b)
y~D,
(c)
Pv(xv;Y)
Let
X'
let
Pv(xv;y)
=
~'
that
any coherent
XI = and
and
on
D.
~I
be
above,
sheaf
Further
Then
for t h e m a p p i n g s
~'
a sheaf
of
p
~xl-modules
~
is the
in
G = D 2 x ... x D m
composite
if and o n l y
if
~I
By i n d u c t i o n ,
by
~.
that
~.(~')
shows
then Let
connected
= O}, of the
to p r o v e
~.'(~')
is
• D m x DIP2(x2,Y) denote
and
x' = p o ~I; to p r o v e
Furthermore,
of
~I
and
then
the
W
(~').(~')
7
is
~x,-modules.
is
of
XI
~ xl-coherent
sufficient
of the
= D 2 • ...
Theorem
Thus we may
it is c l e a r l y
=
of
(~i). (~)
iteration
hence
(~i) . (7')
injection
~G-Coherent.
is
"'" = Pm (xm' y)=O}
=
the n a t u r a l
is a s h e a f
p.(~.(~'))
the
X' ,
~'
Zr D I x W = V be
In v i e w
if
is c o h e r e n t ;
that
D.
"'" = P m ( X m ; Y )
it s u f f i c e s
p.
• D,
~.(~)
on
= XI
and
If
on
it s u f f i c e s
: XI ~ D
so t h a t ~I
the
i.e.
let
~I(X')
p.((~l).(~')),
=
the p r o j e c t i o n . made
: X' ~ X i,
projections.
V,
XnV.
{(x 2 , - - - , x m,y) ~D 2 • ...
let
~~
on
(d)
(c)
coherent
functions
{(x,y) ~ V I P i ( x l ; y)
: X' -~ D
on
= 0 ~ xv~Dv-
= O
remarks for
are h o l o m o r p h i c
above
replace to p r o v e
argument
is c o h e r e n t . • D m • D C Cm + n - I
is
set
Z = { ( x l , x 2 .... , X m , Y ) ~ V l P l ( x l ; y ) = O} and
u
: Z ~ W
discrete above, remarks
fibres
we h a v e (c),
the and only
projection. X'r
Z,
to p r o v e
(f), w e h a v e
Clearly
ulX'
= ~i"
the
theorem
only
to p r o v e
u
is p r o p e r ,
Hence, for the
u.
with
by remark Further,
following
(b) by
statement.
85
(4. i) here
~Z
u , ( ~ Z)
u. (~z)
is a c o h e r e n t
is the structure is, in f a c t , ~
sheaf of
sheaf of for some
by
~W
xI
by
x,
(k = m + n - i),
wl,...,w k
W;
that of
p.
For the proof, we shall de no te (x 2 .... ,Xm,Y) PI (xl ;y) by
Z,
~W-mOdules;
and
p-I P(x,w)
= xp +
a v(w) x v. V =O
(4.I.A) u,(%)
Under
of the functions
qenerate
u, (O z)
Proof of
(4.I.A).
Wo~W
and let
Let
the above hypotheses,
~
over
i, x,
in
c~..~
..., xP-I~F(V,~z)--~ F ( W , u , ( ~ ) )
We have to prove
v I .... ,v r
V.
Then,
in
0 W. the following.
be the points
be a h o l o m o r p h i c
U {vi}
the imaqes
function
there
of
Z
with
in a n e i g h b o u r h o o d
are h o l o m o r p h i c
and a n e i g h b o u r h o o d
Let
N
of
~(vi) of
functions
U {v i}
such that
,W O
p-I (X,W)
=
C v(w) x ~
on
ZnN.
V =O
It is c l e a r l y c ~ V
sufficient
to find h o l o m o r p h i c
and n e i g h b o u r h o o d s W,W
Vi
of
v.
O
functions
such that
1
Iixlon on
p-i (4.2)
Z LV i
Z IV ,
i > i.
l
By the p r e p a r a t i o n P(x,w) where in
x-
Q(x - x l ; w ) s
theorem,
= w(x,w) x
x I (v I = (xl,Wo~),
we can write,
near
Q(x - xl;w)
is a d i s t i n g u i s h e d and
vl,
~
is a unit at
polynomial v I.
= w o-
86
By
w
Lemma
i,
~ W
x,
(4.3)
P(x,w)
for a s u i t a b l e is s m a l l near
Xl;
of
= w(x,w)
Q(x
neighbourhood
WO
enough,
W~Wo,
hence on
Z
vi,
i
vanishes Vi
so t h a t w e h a v e
,w O
'
If
Q
Q(x
these
> 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
d
Q +
(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,woX
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~
then
~
if r a n k
~ O A(x)
if and o n l y if < q.
A(~x)
This p r o v e s
~ ~q
~q,
i.e.
the p r o p o s i t i o n .
88
We that
shall
f
see,
in C h a p t e r
have d i s c r e t e
unnecessary. asserts
VII,
fibres
in the above
A very much deeper
that even T h e o r e m
theorem
7 is true
(with f i b r e s w h i c h m a y be p o s i t i v e
5. M e r o m o r p h i c Let of X.
X
For
(i.e.
the
is not
that the a s s u m p t i o n corollary
of H. G r a u e r t
if
f
is o n l y p r o p e r
dimensional).
space,
~
=~X
the s t r u c t u r e
let ~qa be the ring of q u o t i e n t s of set of formal q u o t i e n t s ~/~, ~ , ~ O a, where
if
~'
carries
a natural
projection
p : ~
~ a , w i t h the i d e n t i f i c a t i o n = ~'~) and let ~ = U ~I The
the p r e s h e a f
a~X
topology X
a sheaf on
a
in
'
quotients
sheaf
a~X,
a zero d i v i s o r
becomes
15
functions
be a c o m p l e x
~/~ = ~'/~' ~
is
such that,
defined
X.
U ~U
This where
of the ring
~
by
p(m)
with = a
set
a"
the n a t u r a l if
ms
)1%
is the t o p o l o g y d e f i n e d b y %
is the c o m p l e t e
of h o l o m o r p h i c
ring of
functions
on
U.
U
It is c l e a r %
can be
that
~/I
set
Every element but,
of
If denote
~
~a
if
a closed can
of ~ a
defines
a meromorphic
not e v e r y m e r o m o r p h i c
the e l e m e n t e
at
~a~a. subset
dense
on a n y o p e n I/~
~U
of
in
U.
set in
can be d e f i n e d
set of poles indeterminacy
of
a.
U.
of ~ ;
consisting of the
of
sheaf
function
on
function
function
on
U
U.
on
comes
a~U
U,
from
the
P
set
x~U - P , x.
a~U,
~
of poles then
at
The
If
d o e s not v a n i s h
of
in the o b v i o u s
U
U, in
the set U.
way.
P~nZ
~
is
and we
U - P
is
identically
t h e n the m e r o m o r p h i c
U - P
a pole
~x~x
set
we
and call
is c a l l e d
of
an
and
defined by
A point
If
I/ ~ in
set of
function
of ~ a
Clearly
speak of the v a l u e
moreover
subring
a~ ~ a . A s e c t i o n is c a l l e d a m e r o m o r p h i c
is a m e r o m o r p h i c
it the g e r m of of
as the
with a subsheaf
o f ~'1,U.
~ by
identified
where
U
in general,
an e l e m e n t
can be
looked upon
the q u o t i e n t s on an o p e n
~
If
function Z
is c a l l e d
is the the
89
PrOposition
7.
The
set of a m e r o m o r p h i c of
set of poles
as w e l l
function
U
on
as the
indeterminacy
is an a n a l y t i c
subset
U.
Proof. of
It s u f f i c e s
~.
If
acU,
to c o n s i d e r
then the g e r m
the
set
~a
of
P = P P
seen to be o b t a i n e d as follows. fa Let ea - ga be any r e p r e s e n t a t i o n
of poles
at
a
is
easily
quotient
of two e l e m e n t s
divisor
of zero.
quotient
as above.
fa' g a r
'
of
~a
as a
where
ga
is not a
Z be the g e r m at a defined by --g,a the set of zeros of the g e r m ga" Then P = N --g,a Z , the D a i n t e r s e c t i o n b e i n g over all r e p r e s e n t a t i o n s of ~a as a
Proposition
8.
Let
This p r o v e s
If
X
is a c o m p l e x m a n i f o l d
at e a c h of its points, function
~
on
X
the p r o p o s i t i o n .
then the
set of poles
is an a n a l y t i c
of d i m e n s i o n
of a m e r o m o r p h i c
set of d i m e n s i o n
n - i
at e a c h of its points. Proof. that
Since X
the r e s u l t
is an o p e n
factorial,
set
is c l e a r l y tic C n.
"
If
fa,g a fl
a~fl.
suppose
Since
~a
is
fa'gar
h a v e no c o m m o n
is small
enough,
holomorphic
functions
f,g
Proposition
i, if
coprime
~b
in
Let
we m a y
we can w r i t e fa ~a - ga'
such that
local,
~
fa,g a in
is small
for any
br
division
ft.
If
Zg = { x ~
I g(x)
x~P
;
then
~xr
fx
in
~x"
But
= 0},
in
by
by Chapter
the g e r m s
We a s s e r t
fb,g b
II, are
n o w that
= 0}.
we c l e a r l y h a v e
so that (fx'gx)
are r e p r e s e n t e d
enough,
Pq) = {xcf2 I g(x)
(except units)
Futher,
n
P~r Zg.
Let n o w
fx/gx,~x,
i.e.
gx
divides
are coprime;
hence
gx
is a u n i t
90
in ~ x '
i.e.
g(x) ~ O,
so that
X~Zg.
ZgC P ,
and our assertion is proved.
Proposition 8 follows
now at once from Chapter III, Proposition Note
This proves that
13.
that we have not used Proposition 7 in the
above proof.
91
C H A P T E R V.
- REAL ANALYTIC
In this c h a p t e r , results
concerning
irreducible complex theory
we
real
shall
which
sets,
following
a satisfactory There
F.
with
substitute
Bruhat
s m o o t h t h e o r y of
is a v a i l a b l e
in the t h e o r y of
for real
importance. analytic we
Such a
sets.
shall
s h o w that
c a n b e o b t a i n e d on c o m p a c t
irreducible
sets of a s p e c i & s
joint paper
fundamental
The
is an a t t e m p t b y B r u h a t - W h i t n e y ,
stating
We
sets.
B r u h a t - H. Cartan,
f a c t o r y t h e o r y of g l o b a l analytic
some of the
is of the u t m o s t
is no l o n g e r a v a i l a b l e
However,
set out
analytic
components,
analytic
SETS
type.
their results;
sets.
at a s a t i s -
components
for real
We s h a l l c o n t e n t o u r s e l v e s
proofs
are to b e
found
in t h e i r
6.
shall
finally give examples
- Cartan
5;
due to C a r t a n
to s h o w h o w p a t h o l o g i c a l
real
II
and
analytic
sets c a n be. W e r e m a r k t h a t the r e s u l t s analytic
subsets
of a n y real
i. The c o m D l e x i f i c a t i o n
in
w 2 would
a p p l y to
analytic manifold without
of a real a n a l y t i c
change.
qerm
Let a ~ R n.
A be a g e r m of r e a l a n a l y t i c set at a p o i n t --a We shall look u p o n Rn as a s u b s e t o f C n. We w r i t e
~a
for the ring of g e r m s of real a n a l y t i c f u n c t i o n s at a~R C for the ring of g e r m s of h o l o m o r p h i c f u n c t i o n s . Let a R of g e r m s of r e a l I(Aa) = IR(_e~)~ d e n o t e the ideal in a
analytic
functions
Proposition
I.
vanishing There
on
n
A..~.
is a u n i q u e g e r m
Aa
of c o m p l e x
a n a l y t i c set at a~C n such t h a t A ) --a A and for w h i c h a n y --a C A Further, g~a w h i c h v a n i s h e s on --aA, v a n i s h e s a l s o on --a" we have
A nR n = A a ,
~a
set w i t h
S ~ --a A --a
the
in
ideal
~C a
and
contains
any
g e r m~
also
S
of
~a
A . --a
If
complex I (-a)~ A
o f g e r m s of h o l o m o r p h i c
analytic
= I C ( -A N)
is
functions which
,
92
v a n i s h on
Aa,
we h a v e I (_Aa) = I (Aa)~RC.
Proof.
Let
fl ....
f '
a~R n
generating
extension
of
= O.
vanishes Rn
that
on
Aa,
g =
ck + ~/~
I(Aa)
then
form
d k.
Hence
g =
ak'~Ca"
ak gk' A
shows
is the
--a
functions gi
I.
A --a
near
be a h o l o m o r p h i c
on
A be --a equations
Further f
if
of
dk)fk
~a"
germ
gi
Let
g
Rn
that
= fi' if
at o n c e
is c a l l e d
of
immediately ~Sa ) --a"
so that
g~I(Aa)_ ,
intersection Aa,
then
dk~aR-
extension
It f o l l o w s
S )A --a --a '
g~a C to
' Ck'
is a h o l o m o r p h i c
It f o l l o w s
sets c o n t a i n i n g
Definition
Aa ) -A- a .
--
~k
analytic
above
analytic
let
(Ck + ~
g = 0
by construction
'
~aR;
the r e s t r i c t i o n
where
our a r g u m e n t
Further
that
f =
~kg k
if a c o m p l e x
Further,
over
It is c l e a r
is of the
Hence
analytic
f. to a n e i g h b o u r h o o d of as n. i a~C n d e f i n e d b y the (holomorphic)
the g e r m at gi(z)
be real r
AanR
n
= Aa;
then
that
of all g e r m s
I(~a)
= I(Aa)~C.
of c o m p l e x
and so is unique.
the c o m p l e x i f i c a t i o n
of the
9 e r m ~a"
Proposition germ) , then
2. A --a
If
A = U --v, A --a a c o m p o n e n t s , and A --a = U --v, a components. Proof.
A is i r r e d u c i b l e (as a real a n a l y t i c --a is i r r e d u c i b l e (as a c o m p l e x a n a l y t i c germ). is the d e c o m p o s i t i o n
of
A
--a
into i r r e d u c i b l e
A is the c o m p l e x i f i c a t i o n of -A- ~ , a , then --~, a is the d e c o m p o s i t i o n of --a A into i r r e d u c i b l e
Suppose
A = --a B u --a C --a
If
A --a
reducible,
i.e.
(B a = --a S nR n, --a C = --a T nRn),
S u --a T . --a = --a
Then
so that e i t h e r
B = --a A --a
93
or
_C a
Sa) --
B --a
= --a A , ,
second
say
--a B = --a A .
we have
S --a
statement
We
shall
analytic
) --a B = --a A ,
follows
write
germ,
Then
dim
so t h a t
at o n c e
d i m R --a A
S
--a B = --a ~ ,
from
for
for
of
= --a A .
S --a
since
The
this.
the
that
so that,
dimension
of
a complex
a real
analytic
germ.
C -a Proposition
3.
dim
A
= dim
A
R --a
Proof. is
We m a y
irreducible.
the
definition
suppose,
because
But
Proposition
If
~c
every
that
then
for
then
of d i m e n s i o n
Corollary. R n,
A
bEU,
.
C --a
is
we
and
a real
point
Proposition
the
has
dim
_~
equation set
If
neighbourhood small If
dimR_~
n
at
analytic if
then
~ dim
A
Cn
b~V,
since
set
inducing
dimc
--bAC ~b'
~b we
a ~ C n.
Then
S --a
is the
A --a
contains
to be
bV
in
an o p e n
set
U
such
. in an o p e n ~a~
and
V
= dim
is
~ - A- a
.
have
an
(which must a germ
irreducible
complexification
then
B --a
complex
be
S a nRn)
dim R ~a and
of
if
m
with
analytic a real
and only
=dimc
~a"
B cS --a --a . S i n c e
S --a
S . --a = --a
~
real
analytic
coherent
functions We
I (Aa)_ = I (Aa) ~ R c -
~ dimc--aA
d i m c--Ba = d i m R Ba = d i m c--Sa'
2.
from
~ dimR--aA.
be
OR-modules. induced
in
S --a
irreducible,
analytic
for
analytic
Let
S nR n --a
said
a
at o n c e
A --a
R --a
4.
Definition is
,
germ
Proof. is
then
= dimc--~
Proposition germ
a complex of
V
enough,
b~VnR
is
S
that
a neighbourhood
R
Proof.
2,
3 follows
analytic
a~
have
of
say
if
the
set sheaf
vanishin~g ' on that
a coherent
A
a germ
A --a
analytic
set.
A
in
~R(A)
an o p e n
set
of g e r m s
is a c o h e r e n t is c o h e r e n t
if
of
sheaf it is
~c
R
real of
n
94
Remark. shall
Not e v e r y real
a complex
coherent
5.
Let
analytic
such that
_Aa
analytic
in an o p e n
equations f..
Sa
=--Aa'
If
for
b~R n
then
since
holomorphic
generate
A --a
Proof.
A 9a . --v
the
ideal
Sb
which
Sb
is the
germ. real
such that I(_~) a9
if
on
for
(where
extension
b~n
so that
~c(S) S
SnR n
induces of
o f g e r m s of
is coherent,
is t h e r e f o r e
A
b y the
the real
s y s t e m of g e n e r a t o r s
irreducible
is
is the c o m p l e x i f i c a t i o n
sheaf
vanishing
A --a
is a h o l o m o r p h i c
S a nRn =--aA,
of
~c(S)
coherent.
components
sets
S = U --v, S --a a = Aa'
inducing
S --v,a 9 b~R n
for
A = S n R n = U (S nR n)
= U A = ~b" --u,b
S --u,b"
near
S b = _~
ideal
a,
S
of a c o h e r e n t
A = U --~,a A be the d e c o m p o s i t i o n of --a A into --a c o m p o n e n t s , and --u~a S = --V,a A the c o m p l e x i f i c a t i o n
Then
dim C (Svn ~ of
the
The
a complexification
_~
generate
p a r t s of a finite
6.
Let
analytic
are
a,
acU,
fl .... 'fr'
a~,
near
C n, Then
analytic
gi
and
are coherent.
irreducible
where
as we
germ,
in
~r R n,
that
then
~R(SnRn),
Proposition
U
near
and if
where
functions
and i m a g i n a r y
b~UnR n
is defined,
It is t r i v i a l
SbnRn,
of
S
= O,
If
germ
set
(fi)b
gi(z)
analytic
set
of some real
then
--a
for
is coherent 9
the g e r m s A )9
if,
_~
If
induces
be a real
is its c o m p l e x i f i c a t i o n .
if and o n l y
Proof.
b~,
A --a
set in an o p e n
S = --a A --a
complexification
A . --a
set is c o h e r e n t
see later.
Proposition
of
analytic
S~) b
=
and
Furthermore
9
if
S
are V
S = U S ,
then
V
a.
U Av
say. C l e a r l y
if
b
Hence
is near
enough
of any i r r e d u c i b l e
we have,
obviously,
~
S. --D
is
~b = ~/~"
near
Furthermore < dimension
so that
p,
br
to
a,
component b"
95
Hence
any
irreducible
> d i m C (A_>,bn~v , b ) it f o l l o w s
that
equality,and
A --a
if
7.
of
S has d i m e n s i o n --u,b Since S = U (~ , bnSv,b ) ' --u,b
~ ~ v.
--u,bSc ~ v , b ;
A --v,a
Proposition
component
~v, ibc ~u ,b '
since
we h a v e
is c o h e r e n t .
If
is i r r e d u c i b l e ,
A
is a c o h e r e n t
then
for
b
real
analytic
a,
we h a v e
near
set
and
d i m R--~ = d i m R ~ a. Proof. for
If
b
S
near
is a c o m p l e x a,
we h a v e
~b
~_~ = d i m c S
d i m R--~ = d i m c
8.
Let
with
R 2n
z3 = x3
+ lyj" .
Consider
the g e r m
A --a
is c o h e r e n t
complex germ
~b
= d i m R--a A .
set
if and o n l y
S
(We h a v e Proposition
(xl,. - - , X n ; Y ~• , "'" , yn) ,
analytic if,
at
for
b
set b
is
A.
near
Then a,
again
irreducible. Proof. by
Consider
~(z i ..... Zn;
u 3 = z 3 + i~j, complex
if
= {z~cml
sets
~i .... ,In)
=
map
2n
~:C
(u I ..... Un;
v 3 = z 3 - i t j,-
E
~E}
an a n a l y t i c in o p e n
irreducible El • E2
C-linear
clearly
2n
~C
defined
v i .... ,v n)
~ I R2n (=C n)
where is the
conjugation.
Now,
again
the
sets
is a c o m p l e x (~
is the c o m p l e x
set. in
at the p o i n t s
IV,
set
in
m C ,
conjugate),
Theorem
a I, a 2
at
a =
respectively,
(a I, a2)
l, C o r o l l a r y
6).
E
is
analytic are
then
(as follows, Hence
and
then
Furthermore, if El, E 2 are mI m2 C , C respectively, which
is i r r e d u c i b l e
from Chapter
analytic
2.)
analytic n Identify C
S.
as a real
induced by
then
complex
(z I ..... z n) ~ S
= --a' A
Hence
be an i r r e d u c i b l e
b y the m a p p i n g
~a
is i r r e d u c i b l e ) b y
induced by a complex analytic
germ
the
= --~"
S --a
S --a
set w i t h
b -~ d i m c ~ a
S ) since d i m C --Sb = d i m C --a Proposition
analytic
e.g.
96
-i
(S x 3)
= X
a(~C n = R2nc
is a c o m p l e x
C 2n)
= dim R ~ a "
dim c ~ a
complexification Suppose ~b
of
that
of c o m p l e x
irreducible
at
X ) ~a and --a _X a is the
4,
A . --a
for
some
where
analytic
b
arbitrarily
neither
~ -I (~•
above,
set,
we have
By P r o p o s i t i o n
= ~ l , b u~2,b'
remarks
analytic
and c l e a r l y
~i,b
x S--l , b ) u - i
is
near
a,
~b"
By our
(~2,b • ~2 ,b )
set c o n t a i n i n g
we have
is a germ
_~,
and
but b y our
this
implies
that
&
that
if
is reducible,
a s s u m p t i o n on the S. b' -I ---l, ~ _~ = ~ (S__ b • S b). This shows
then
of d i m e n s i o n that
A --a
= dim R--~'
This
components
section
of i r r e d u c i b l e We shall theory more
_~
= _~,
state
to F.
disadvantage
of a real
components
also
due
is irreducible, analytic
which
Bruhat
of sets.
Lemma
i.
Let
analytic
germ
implies
such
components,
~ O, A that
of the t h e o r y
and H. Cartan.
This
global,
latter
open
set in
let
each of w h i c h
has o n l y contains
the
restricted
R
A = {x~
a fundamental
U - A
is
but has
o n l y to a m o r e
and
has
set
of an a l t e r n a t i v e
it is more
applicable
of
Bruhat
results
be a c o n n e c t e d
a U
to F.
and H. Whitney.
function,
e v e r y point
neighbourhoods
due
in that
~
analytic
to an e x p o s i t i o n
the m a i n
of being
class
connected
so
is d e v o t e d
satisfactory
a real
and
&;
is coherent.
2. I r r e d u c i b l e
Then
~b
A is not coherent. Conversely, if ~ b --~1 ~ (S_b x ~ b ) is an i r r e d u c i b l e c o m p l e x
then
so c o n t a i n s
n
and I f(x)
system
f = O}.
of
finitely the point
many a
in its closure. Proof. the
lemma
The
lemma
already
is trivial proved
for
n = i;
for analytic
sets
we
shall n-I in R
suppose
97
By t h e
preparation
theorem,
we
may
suppose
that
a = 0
and
P that
f = xp + n
polynomial. factors, of
f
hood has
is ~ O.
only
which
is I
x' ~U' , Let U - B
is
adherent
to
only
Let
- A'
x'~U'
O
Xn(I,
R
= O})
such U
sufficient
that
= U' to
/ 0
roots
of
p~
in the
• I.
prove
components,
6(x')
integer
neighbour-
U', e a c h of v in its c l o s u r e ) .
let
connected
real
is an
in
= 6(x')
{x'r
O
and
is
Since of
=
contains
about
many
number
(A'
small)
components
It
no multiple
that
each
U'
of
and
v'
U'
v
polynomial
independent
of
x'
n
rI
< r2
rp
+l(x') < j
< ...
< rpv,
= +c.
~ p
.
continuous
are
obviously U proves
Clearly
U~.
the
sets
now
For
any
=
v,j
A
these
if
I =
r3 (x')
it is c l e a r
on
the
Let
be
and
Further,
are
R n.
U'
= O}uA.
O.
it h a s
(arbitrary
(i.e.
rl(x') .... ,rpv(X')
This
that
an
imply
to
is a d i s t i n g u i s h e d
6 ( x I ..... X n _ I)
connected
finitely
the
for
that
O
= O
is a d h e r e n t
x )
be
interval
{xEUl6(x')
has
'
O
many
connected,
f(x'
U'
such
f(x' ,x n)
B =
which
Let
an o p e n
suppose
discriminant
finitely
be
.. X n _ i ) x p-v ' n
'"
we m a y
the
O~R n-I
(x I
~
Further,
so t h a t
of
Let
a v=l
The U
roots
(-c,c),
O
{x'~U'v,
r
3
suppose let
as
x' -~ O
that
the
functions
components
~ j .~ p~,
(x')
< x
n
that
ro(X')
~ 0
connected
v,j'
and
= -c,
if rj
of
U
- B
where
< r
j+l
(x') }.
lemma. be
an
integer
analytic p,
O
{ p
set ~ n,
in an o p e n we
denote
set
Q
by
R
in (A) P
the
set
of
regular
Proposition in Then
Rn
9.
which a
has
points
Let is
A
of
be
irreducible
a fundamental
A
an at
of d i m e n s i o n
analytic a point
system
set
p.
in
aEA.
an o p e n Let
of n e i g h b o u r h o o d s
set
d i m --a A = p. U
such
98
that
UnR
(A) has o n l y f i n i t e l y m a n y P each of w h i c h is a d h e r e n t to a.
connected
Proof 9
After
We m a y
of c o o r d i n a t e s be
such that
Lemma
in
the
4 apply
By C h a p t e r
suppose
that
Rn
let
assertion
U'r
R p,
U"r
of C h a p t e r
to the g e r m
III,
a = O.
A_o.
Proposition
Let
5,
W
is dense
P in
P This follows map
f
U = U'
Propositions
2, 3, 4,
Wp = {x~AnUI6(x') r R
x U"
(A).
~ O}.
We c l a i m
that
P
R
from the
fact
of a m a n i f o l d
fibres noted
change
(A)nU = V (say). In fact, no open set P P can be m a p p e d b y the p r o j e c t i o n ~ into {x'~U'16(x')
V
in
a linear
R n-p,
III,
P
W
components,
of
f
into
are discrete,
already
to prove
X
that
in C h a p t e r
that
W
if there
a manifold
then
III).
has o n l y
exists Y,
an a n a l y t i c
such that
dim X ~ dim Y
It is t h e r e f o r e
finitely
many
= 0}.
the
(a fact sufficient
connected
components,
P each of w h i c h by Lemma
I, that
connected in
and
Let
U',
~k(j) (x')
~ O}
further
has o n l y
each of w h i c h < 9 ..
(x')
~J)
x' -* O.
x')
(Pu
Further,
finitely
< rpu(X')
= O,
suppose,
is a d h e r e n t
many
to
O
be the real
is c o n s t a n t
= Qj(rk(x'),x')/~(x'). (J) Pj ( ~k
as
We can
P(Xp+l;x')
Lemmal 4, we have (x') -* O
O.
< r2(x')
rl (x')
of the p o l y n o m i a l
let
to
{x'~U'16(x')
components
R p.
roots
is a d h e r e n t
Then,
on
U~),
by Chapter
III,
so that
if (j)
W
u,k
then
= {xEAnUIx'~U',
W
v,k proposition
Theorem R n.
I.
Then
are the c o n n e c t e d
components
of
=
~k
(x'),
j >~2},
and the
W , p
follows.
Let every
A
be an analytic
point
f_ollowinq
~ro~erty:
qerms
) ~a'
~a
Xp+ I = rk(x') , Xp+j
a~A
i_~f S
then
has
set in an open
a neiqhbourhood
is an analytic
SnV) AnV.
set in
set
V
with ~
~
i_nn the
and the
99
Proof. of
We
A.
that
shall
(If A --a
p = O,
set
decomposition
the
V'r of
~,
S
in
V'
A --a
take
with
V =
of
in
a
suppose
and
that
that
a nonempty
in the
Corollary
I.
analytic
sets
stationary
the
only
if set
functions
O,
a
for
~
Corollary
many O.
then
Now, SnX
Xv
is a d h e r e n t
Xv
S) A ' n V .
S
is
This
V
has
I.
the
be
{As}
~ s o.
then
of T h e o r e m for
A a) A ~ o n V , i follows.
family
family
of
is
~.
the
ideal
on
-A- ~ , a
in .
V
be
n,a Let so
be
filtered,
a neighbourhood
I relative a
of g e r m s
0
is d e c r e a s i n g l y Let
O) ;
there
if
that
set
IS
to
By i n d u c t i o n ,
in T h e o r e m
in
is an
containing
filtered
set
if
V
such
~r R n,
= O},
connected
is a d e c r e a s i n g l y
I s = I~o
Hence
then
~ p - I
of d i m e n s i o n
to
S_o) A_o,
V r U,
vanishing
Since
manifold
finitely
(since
a neighbourhood
(x~AnUl6(x')
manifold
required
and
if
set
is a d h e r e n t
and
~ a ) ~a'
in an o p e n
is m a x i m a l ,
> 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
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;
A
in
of
S
Further,
~, A r S.
is the
B ) K ( S - A).
exists. A).
subsets
By
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
Corollaries
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
relating
Whitney
long.
= 0
is a c o h e r e n t / ~R},
A
to, b u t
9 above.
analytic
functions
for all
i}.
s h e a f of i d e a l s
i.e.
6.
They depend
similar
A subset
finitely many
~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 and
irregular
follows,
y,
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
of
term
on
the
This
proves
that
S
vanishinq
on
identically Example by
3.
that S
can
that
where is n e a r
il)
such
S
set.
right
H. C a r t a n
One
analytic
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)
are dense
and
for e a c h
T@ = i d e n t i t y . that
integers,
TI(D) = T (Dk), 2 ~ x + y + z ~ n(I)
S =
irreducible
then
TI(A')
Let
of p o s i t i v e
of e l e m e n t s
A'.
is
and
sequence
an i n f i n i t e
TI(A')
then
k,
set
R 3.
any
the
family
finite
of
not meet
A I ~C,
many
P,
= x3
the number
are d i s t i n c t .
locally
be
contained
I = {J, k}, TI(A')
subset
Choose
motion
is
the
points
{x = y = O},
(n I ..... np)
a euclidean
(so t h a t
set c o n -
is an o p e n s u b s e t o f A. T h e r e is a 3 of R leaving D invariant which takes
k = 1,2 ....
that
line
any compact
p = p(I)
n = n(I)
set
D - {0}
euclidean A'
the
of
A).
z ( x 2 + y2)
and
is
Sn - P
the TI(A')
is a n a l y t i c . if Proof.
are p r o p e r l y infinitely many
k,
AI, ~ C
If
108
where
I' = {I, k}.
A{I,k}r S'
for t h e s e
= U Aj,
p(J)
J
S"
= U Aj,
or
p(J')
over
whole
sets.
of
B
there
irreducible
(p(I)
b e o n e of them. finite
that
t e r m of
S'
exist
analytic
p(J)
S"
nor
J'
are
is the
B
sets which
is
again
S"nB
Hence
Let
~ p(I)
t e r m of
and
(S'nB) u ( S " n B ) .
for w h i c h
j = k o.
either
S'nB
Let
sequences
+ I) th
neither
B =
is i r r e d u c i b l e
for w h i c h
the
Clearly
k~
+ I) th
J'
shown
A
those
(p(I)
and
and
Hence,
Let
those
It is e a s i l y
since
over
or t h e
> p(I)
analytic
k.
running
~ p(I) ,
k o.
Hence,
is r e d u c i b l e .
have
no
components. 3
Example given
A
Let
5.
be
the
irreducible
analytic
x 2 (z + I) 2 + y2 (z - I)2 =
(z - I) 2
set
R
in
by
contains
the
it c o n t a i n s D',
A
whose
lines
further union
D = {x = O,
a family
A'
meets
of
z = i},
D'
= {y = O,
lines meeting
D, D'
both
in a c o m p a c t
D
z = -I}; and
set
(lyl ~ 2, Ixl ~ 2 r e s p e c t i v e l y ) . If K is the c i r c l e 2 2 x + y = i, z = O, then A = DuD'uA'uK. Let
I
positive
run over
integers
arbitrary). by
p.
I = {J, J',
elements. Dk
on
Take
(resp.
coplanar,
D~,
TI(D') x
2
+ y
k},
k = I,
+ z
2
Let
while ,< n(I)
finite
p = O,
sequences
elements
transformations
where
and are dense
= D~, 2
For
of all
2P - I
affine
on Tj(A')
Tj(A') , Tj, (A') .
family
containing
We define
induction
then
the
T~
(p) 0 TI
identity.
J, J'
contain
as f o l l o w s , If
2p-i
Tj(A'))
a sequence
2 .... ),
so t h a t
D k,
TI
be
TI(A'uK) In(I)
family such does
D k'
of p a i r s
that
TI(D)
not meet
= sum of elements
p > O, - I
resp.
in the
of
of
lines
are n o t
of
lines
on
= D k,
the
set
in
I).
This
109
latter DuD' S
condition
can be ensured:
in a c o m p a c t
is an a n a l y t i c
Proof.
Suppose If
having
2p - i
for
(since
p
and the result
having
the
AI~ ~B'.
But then,
A H ~BuB'
dimension
any set
I.
S
2
contains
= S,
Hence,
~DkC
contain
the w h o l e
Thus,
there
AI) 9
then
I
with
there
is
11
then
~B
TI(A')
is
k
with find
that
there
D k, If,
AII
is a
|
I1,
~B, k
for
z I, k}, a c o n t r a d i c t i o n . many
analytic
the line
set c o n t a i n i n g hence and
is irreducible.
to prove
on each
I),
S = U A I-
subsets
singular
set of
D k c TI(A') this
for
singular
contains
it follows
of
TI(A')
that
it m u s t
S.
exists
an i r r e d u c i b l e
2 containinq
of d i m e n s i o n
2, a n d
the
is
Further,
of density),
of d i m e n s i o n
taininq
infinitely
segments
of
there
such
H = {If,
(for any
(by our a s s u m p t i o n
S
we can
where
meets
It is e n o u g h
Hence,
as above,
any analytic
TI(A' )
Then
there
arguing
the
nonempty
hence
of elements,
contains
(viz.
A I ~ B,
follows.
same n u m b e r
Clearly
S
If
is irreducible),
D~ ~B
which
AII ~ B.
A'
and let
2P - I >p(I),
with
+ i,
2.
B ~ S. Then
is so that terms
that
A I = TI(A),
set of d i m e n s i o n
p = p(I) A
Let
S = BuB',
A I ~ B.
this
set.
note
infinitely
such that
set of sinqular
the
points
analytic
many
smallest of
S
set
analytic analytic i__ss S
ScR
3
subsets set con-
itself.
110
CHAPTER
VI.
The
- THE N O R M A L I Z A T I O N
aim of this
fundamental in m a n y but
theorem
contexts.
proves
X'
such
f
of regular
flUnX'
on
have
~aC~a a
which
of the germ of Chapter defined
~a III),
by
h ~ u.h
ideal
in
~a.
Lemma
I.
~a
Proposition
I.
Proof.
We have
element
fg'
ring
n C ,
if
a.
irreducible
exists
because
we
component
of the results ~a ~ 0 a
and maps
~a
onto
an
we o b t a i n which
is c o n t a i n e d
of the ring
integral
a,
denominator
on any
of q u o t i e n t s
is the
X,a X at
is a u n i v e r s a l
module
These
=~
a on
U
we m a y define
at
O
functions
~a
in
holomorphic
aEX,
is noetherian,
is the
~
a neighbourhood
function
is i n j e c t i v e
Oa
ring
has
weakly
~a-homomorphism
is a finite
complete
of q u o t i e n t s
u
~a
X.
For any
u
the
Since
in the c o m p l e t e
Let
then
of
Obviously,
(such a
to
A holomorphic
holomorphic
not v a n i s h
due
17).
is c a l l e d
if
proof
on a c o m p l e x s ~ a c e
aEX - X'
Further,
does
see
b y themselves.
points
of h o l o m o r p h i c
.
complicated,
set
X'
Oa.
is basic
interesting simple
of a
set in an open
is bounded.
form a ring of germs
are
functions
defined
ring
at
which
which
is rather
(unpublished;
the germ of a w e a k l y germs
proof
two proofs
27
a beautifully
if e v e r y point that
is to give K. Oka
be an analytic
the,set
function X
ends w i t h
X
to
first
holomorphic
Let
on
The
and R e m m e r t
I. W e a k l y
and
due
some results
The chapter Grauert
chapter
THEOREM
closure
~a"
of
~a
in its
of quotients.
of
seen
that
~a
is c o n t a i n e d
in the ring
Oa"
where
of the ring
g
is n o n - z e r o
of q u o t i e n t s
of
divisor ~a
in
which
~ a'
be an
is integral
111
over
~ a;
there
exist
h i ..... hp~C9a
(1.1) at
+ hl,g"
a.
Hence,
f, g, h i
there
dense
+
... + hp = O,
is a n e i g h b o u r h o o d
are h o l o m o r p h i c ,
is n o w h e r e
so that
in
X,
such that such that
U
of
the set
a
in w h i c h
{x~UIg(x)
= O}
h i are b o u n d e d on f U and (I.I.) h o l d s in U. But then -- is a h o l o m o r p h i c g f u n c t i o n on {x~Ulg(x) ~ O} w h i c h is b o u n d e d because of (I.I)
and so is h o l o m o r p h i c
Chapter
I, P r o p o s i t i o n
Conversely, that
~
~a"
is i n t e g r a l
over
Remark.
and b o u n d e d f Hence gr
ii.
let
From Chapter
the
on
Since
UnX'
by
Lemma
i implies
0a.
III,
Theorem
6, C o r o l l a r y
~a is n o e t h e r i a n , it follows is the ideal in ~ a of g e r m s at a
3, and the
fact that
that
~a
of h o l o m o r p h i c
functions
vanishing
is an i n t e g e r
Proposition
k
2.
on the s i n g u l a r set of k~ ~ 1 such that ~aOa r O a.
If
X
then for any b o u n d e d X'
is a c o m p l e x
holomorphic
is the set of r e g u l a r
space
function
points
and
U
if
X,
and
then there
and h
a~X
X__a on
irreducible, U n X',
where
is a n e i g h b o u r h o o d
of
a,
lim h (x) x~a x~VnX' exists;
here
V
runs over the
Let
E
be the
Proof.
a sequence
xu
Proposition
i above,
a, x u ~X' ' E
we have o n l y to prove be a sequence such that ~a
set of
is irreducible,
~C
of n e i g h b o u r h o o d s for w h i c h
such that is a finite
that
E
of n e i g h b o u r h o o d s
V nX'
filter
is c o n n e c t e d by Chapter
lim h(x~) set.
a,
= ~.
To prove
is connected. of
there
Now,
a.
exists By
Proposition let
{V~}
V v + 1 r V u, N V u =
(such a s e q u e n c e
exists
III,
ii).
Proposition
of
{a}, since
Clearly
2,
112
E
where
Ku
is
V vnx'
is c o n n e c t e d ,
decreasing
3.
singular have
compact This
proves
the
--
component
holomorphic that
that
in
Cn
that
Let
the
coordinates
h
so t h a t
proposition.
X_a
is
in
the
of Chapter
is a p r o p e r
map
with
Proposition
5,
~(S
= dim
in C h a p t e r
there
finite - S')
- B'
III,
III
analytic
are
and
set
fibres, is
is p r o p e r , 5,
that
so t h a t
Now
- B'
such
since
that
dim
Proposition
12,
B'
U',
so that,
on
U'
Hence
weakly
dense
{ dim
that
to
set b e i n g
~(x)
f in
holomorphic.
+ U'
in an o p e n a = O for
av
U'
is
if
f
is b o u n d e d S,
f
it
admit small on
a suitable
p=dimS
Then
~
, --o : SnU ~
by Chapter
finite
is h o l o m o r p h i c aI
...
follows
SnU
is b o u n d e d
- B, on
Hence,
on a
on
= O
on
SnU
from Chapter
they and S',
of
' m
holomorphic enough,
U'
IV, U'
fibres.
fm-V(x) av(x(x))
the
as
a~X,
a n a l y t i c set in -i If B = x (B'),
functions
- 2,
any
at
'
U'
set o f
an
with
if
holomorphic
for
that
verified.
~ p - 2.
Theorem
are
that
holomorphic
so c h o s e n
= B'
the
X.
irreducible be
A
• U" , U' r C p , U" r C n-p,
(S - S')
- B ~ U'
on
is an
Cn
U = U'
space,
Suppose
is w e a k l y
X = S
neighbourhood conditions
a complex
sets
a
~ dim X - 2, where X is an --va -~ a of X . Then any holomorphic function --a
suppose
- B,
connected
is
( X v a n A a)
set
SnU
{K v}
= X - A.
To p r o v e
as
Thus
Since
X'
Proof.
: SnU
Kv.
h(VvnX') .
points,
is w e a k l y
dimension
set
be
X'
we may
of
Let
--
on
the
X
dim
irreducible
of
so is
is c o n n e c t e d .
Proposition
we
closure
sequence
E = N Kv
its
the
u01 K v
=
III,
extensions are this hence
bounded latter is
- B.
113
Theorem i.
Let
X
be a complex space such that,
denotes the set of singular points of dim
(~a n ~a ) ~ dim Y_a - 2
component of
~a"
holomorphic Proof.
X
in an open set point of morphic
Let U
U, while
~
X
being any irreducible
of germs of weakly
is an analytic
fl ..... fk
of
~
A
we have
is a coherent sheaf of
We may suppose that C n
a~X,
Then the sheaf
functions on
open set in
of
for
X,
if
~X-mOdules.
set in an
be holomorphic
functions
X which are universal denominators {xEUIfl(x)=...=fk(X)=O}=AnU.
function on
Let
u be a holo-
U which is a universal denominator
U, which does not vanish on any open set in
U.
at any
at any }oint
These
functions
exist by Chapter III, Theorem 6, Corollary 2. Consider the sheaf universal denominator,
~=
--~TT over
prove that
is coherent.
(1.2)
~U"
(1.2).
Suppose
(fi)a f = (u) a"
Hence
It is therefore
(fi)a fs (u) 2 a ' a
fi
sufficient to
_r u
Proposition
since '
U
if and onl~
f
is holomorphic fi (x) ~ O.
being
1
Conversely,
if
f~a'
at any point
x
Since the common
form precisely the singular
and the singular set of
has codimension(
set of 2
U,
at any point,
3 implies that
h =--f~(ga; hence u This proves (1.2).
f = (U)a h~u. ~ a = ~ a" Now,
(fi)a h ~ a .
at which at least one
zeros of the
which is
f = (u) a" h, h ~ a .
Then
(fi) a h~(u) a" ~ a '
then
is a
We assert
f~.
a universal denominator,
u
~T;
f ~ a satisfies f ~ a , aeU, i = 1 ..... k, (fi)af~ (U)a~a .
Proof of
near
Since
An element
if, for
and
U.
this is a subsheaf of
isomorphic to J~
u. ~ U o n
for fixed
i,
the set of
(fi) af~ (u)a~ a form a coherent
f~Qa
subsheaf of
sheaf of relations between the sections
such that QU fi' u
(since the of
~U
is
of finite type). The intersection of finitely m a n y coherent subsheaves of
~U
being coherent,
(1.2)
implies Theorem i.
114
2. N o r m a l
complex
Definition.
spaces
Let
X
normal
inteqrally said
said
to b e
to b e Note
be
a complex
at
a
if
closed
in
its
complete
normal
if
it
i_s n o r m a l
that
if
Remark.
Let
X
nowhere
dense
holomorphic U
map
so t h a t the
- A
= f.
In
closure
of
functions; that
2.
of
is
: Y ~ X
X
a normal such
has
because
is
the
ring
to
say
that
its
X --a and
X
is
i__ss
points. irreducible.
A cX k C be
a a
a neighbourhood f
admits
of
is
map
a holomorph~ k F : X ~ C
Proposition
of germs X
1,
of w e a k l y
is n o r m a l
function
at
a
at
a
is
easily.
a complex
the
o n e L of
a holomorphic
complex
that
,a
: X - A ~
Then
follows
be
%=%
then
f
holomorphic
assertion
Let
X
hence
X
space
a(A
fact,
any weakly
Our
is
~a
every a,
is b o u n d e d .
FIX
Definition
~
- A
as
of q u o t i e n t s .
complex
every
there
holomorphic.
map
that
i.e.
say
(Y,~)
such
at
Let
and
rinq
rinq
at
subset.
X,
integral
to
a normal
to
holomorphic is
be
flU
local
is n o r m a l
analytic
for w h i c h
extension
X
the
space
space.
space
followinq
A normalization
Y
and
a holomorphic
conditions
are
satisfied. (a)
: Y ~ X
(b) A an
= ~
-I
i_~f
(S),
then
analytic
Lemma
2.
there
is
S
is Y-A
isomorphism If
an
(Yl,xl),
analytic
is the
proper
and
set
sinqular
of
is d e n s e onto
in
has
Y
finite
fibres;
points
and
~
of
IY
X
and
- A
is
X - S.
(Y2,x2)
isomorphism
are ~
normalizations
: YI ~
Y2
of
such
X,
that
~1 = ~2 ~ Proof.
Let -I = ~i (S) ,
S
be
Ai
i = I,
the 2.
set Let
of
singular ~'
: Y1
points
- A1 ~
Y2
of
X,
- A2
and be
115
the analytic a holomorphic Proof. This of
Let
-I ~2 ~
isomorphism map
We assert
~ : YI ~ Y2
Yo~Al
and
such that
Xo = ~l(Yo) "
that
there
is
~ I YI - A i = ~'.
Consider
~l(xo)
.
is a finite set. There is t h e r e f o r e a n e i g h b o u r h o o d U2 -I ~2 (Xo) isomorphic (by a map ~ : U 2 ~ S 2) to an analytic
set
S2
in an open
hood
of
~21(Xo )
exists
since
set in
which
~2
C n.
Let
is s a t u r a t e d
is proper).
Let
V 2 r 1 6 22 with
V
be a n e i g h b o u r -
respect
~2
(which
be a n e i g h b o u r h o o d
of
a n
such that This m a p (since by
~I(V) r ~2(V2), admits
Y
~(x)
a holomorphic
in
~
-I
same way,
on
which
=
hence
Let
2.
~(V 2) r S 2 r C
~ : V ~ ~(V 2) r S 2
follows
easily
is a h o l o m o r p h i c
(~,)-I
on
X
is normal.
X
there
YI - AI'
Y2 - A2'
Theorem
: V-A I ~
extension
Our a s s e r t i o n
I Y2 - A2
= identity on
~o~'
(since
YI - AI
YI ) .
In the with
let
is normal); let ~ : V ~ Y2 be the m a p d e f i n e d -I = ~ o~(x). Clearly ~ is h o l o m o r p h i c and
~ I V - A I = ~'. is dense
and
Y2"
: Y2 ~ YI -I
= ~I-I ~
hence
-1
map
Clearly,
on
YI'
and
~o~
~ -I
This
proves
Lemma
be a c o m p l e x
space,
and
Then,
is the
sinqular
if
S
o~ = identity
2.
aEX
a point set of
at X,
we have dim --a S ~ dim --a X - 2. Proof.
Since
-X- a
a neighbourhood Suppose
that
containing ideal
is i r r e d u c i b l e
U
of
there a
on
such that R = ~a
(SI) a.
that
~
contains
fact
~
would
such that
is an i r r e d u c i b l e
in the ring
vanishing
a
if
Then
no prime
define
p = dim ~a' dim ~
component
dim S I = p - I. of germs ~
~
for Si
Let
~
of h o l o m o r p h i c
is a prime
ideal
= p
of
a g e r m of a n a l y t i c
ideal. R set
there
is
bcU.
of
UnS,
be the functions
We assert
properly;
in
--a'Y and we
116
would have
(Sl)a ~ ~a ~ ~a'
which is not possible
(by
Chapter III, Proposition 7) since
dim Sla = p - i.
be the localization of
viz. the set of quotients
a/b, a, b~R, b ~ . R' = ~'
R
at ~ ,
say. Further since
~
X
is normal at
contains no proper prime ideal in
is maximal in is
R' ,
a,
R
hence the ideals of prime ideals. Let
n > O, n~z Since
u~'
principal
R'
R'
u~(~') 2
which
~'
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'
exists because
Then the principal ideal
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
set in
let
cn.
V
X - Y
points
is small
h(x)
= 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
V
clearly
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
(~)
I =
~ z
g = i
an e l e m e n t
Ik z l (v)
=
'"
on of
.. ' z(V)) m
'
it is clear
that
if
V~,
N / v.
= 0
on
he nc e a' we have 0 =
(a, z (v))
of
a
projects
into
an ele m e n t of
~a"
this m e a n s
that
(y) = x
near
This proves
(3.1)
Y
near
v~, by P r o p o s i t i o n
a, = 0
on
uihi,
X
so that it T h e o r e m
be any c o m p l e x
sheaf of qerms of w e a k l y h o l o m o r p h i c
Proof. seen that
Let ~X
sheaf of
(Y, ~
D
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.
2,
and is a n e i g h b o u r h o o d
near
Yo'
and w i t h
Let
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
ui(~(y) ). zi, if y = (x,z I ..... Zm), i=l (u) Yo = (a, z ) on Y. Hence ~ is
on
4'
Hence m
Then
~ =
holomorphic
is
uihi,
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
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
consider
v / ~.
on
regular
OX
V v,
Further,
Vv
v
(x,z (v))~Y.
In fact,
g =
that any s u f f i c i e n t l y
of
Hence
ui(a) z(U) i '
z (~) ~ z (u)
V
Proposition 2,
v.
for each
(v)
u i(a) z i(N) '
w h i c h proves
Theorem
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 is a finite ~ - m o d u l e (containing since x x N x 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
Y
that
for
aEA.
dim
Y
any A
the set
Y.
the
analytic
subset
of dimension
local
analytic
Z
be
is
sufficient
is
analytic
Y,IX'
the
set
to b e
Again,
if
with
~
Let of
A =
~ - Y
is t h e u n i o n m).
of
It
dim
an
analytic X,
X'
of
of
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,
means
on
a~Y,
an
Y
X - Y.
is n o n - s i n g u l a r ,
a neighbourhood
i < p.
(Am
Suppose,
all
of
is
~c C n.
X'
intersection
dimension
b~A m
of
i__nn X is
It
is n o n - s i n g u l a r .
A
and
~ -I
theorem
for
Since
automorphism
of
Let
in
hypothesis).
subset
such
proved
A
assumed
a comple x space
an a n a l y t i c
= ~ - Z.
of
be
p ~O
an o p e n
< dim
are
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
x A,
and
A
We g i v e o f the
The
theorem
complex exist
U'
- Z)
is an
of
theorem
Let
Hence,
by
the c l o s u r e
A,
in
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
a
is p r o p e r ;
III
in
x, h e n c e
~(y) ~Y'
U),
X
U.
point
3 to T h e o r e m
< ~},
is p r o p e r .
Z ) Y.
is d e n s e
isolated
of
with
~ i < m.
is a n a l y t i c
immediately
above
of
by construction,
with
the
intersection
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 on
< 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
Pl ..... P k ~ C Z o
many
Any
space
analytic
~N
such
A
is an a l g e b r a i c
homoqeneous
..... 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
Cz o .... ,ZN
many homogeneous
vi
is N o e t h e r i a n ,
polynomials
i = 1 .... m Ikl " '
v, i = l t . . . # m .
there e x i s t
Pl ..... P k ~ C Z o ' ' ' ' ' Z N
,< 1.
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
25
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
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
=
by
such
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).
the r e s u l t
theorem
map
being
that
that
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
theorem.
clearly
P.
in
that
Mk C
dimension
this
C n.
k-
(see e.g.
Let
Proposition
to
5.
is an analytic
f
then
P nU.
Hence
Pf = @
has
(by
a holomorphic in
U
(and
extension
without
using
the g e n e r a l
a direct
proof
is a n o w h e r e
case in the
functions 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"
{(x,fI(x) .... ,fk(x)) I x,X - P}. isomorphic
of
so
dependence
set of poles
Let
4,
theorem
and the c o m p l e m e n t I.
ai~P f,
unless
However,
be the p r o d u c t cMk
U - Y
theorem.
be m e r o m o r p h i c
P
in
is a m e r o m o r p h i c
the
long
the
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
F/g k
and a l g e b r a i c
and
Further
component
gA
to prove
fl ..... fk
of
Proposition
Since
is rather
of poles
8).
proves
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 2,
to those used
functions
is an alqebraic 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
14.
I I0
In less detail,
I.
in Herv~
19
in
theorem, proof to
II.
due to
sets over a r b i t r a r y
Theorem
J.P.
Serre.
theorem
I,
stated on page
i, which is one C. }{ouzel in
12
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.
in
stated here w i t h o u t
For an analogue of the p r e p a r a t i o n
Chapter
(mostly a l g e b r a i c a l l y
of Houzel
as stated here,
is proved by
preparation
is
18.
The rank theorem, Chapter
That in A b h y a n k a r
(and/or A b h y a n k a r
F. Hartogs'
-
in algebraic geometry.
Most of the results
14).
is proved
work
in Gunning
although v e r y d i f f e r e n t
fields
The ideas in the treatment
are proved Rossi
in Herv6,
non-discrete
from G r o t h e n d i e e k ' s
18-21 by
it is treated
and treats analytic
c o m p l e t e l y valuated,
Chapter
Exposes
is based on the same ideas.
quite d i f f e r e n t
drawn
and 1960/61,
The treatment
in detail,
closed).
and in the seminar notes of
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,
although Oka
and Cartan h i m s e l f
that he understands
Oka's version,
from that of Cartan,
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
6.
v e r y different
proof,
16, 15.
and is These
from the one we have
quite different
This chapter
ii,
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
The unproved results concerning
in Cartan Whitney
II 6
(Proposition (Proposition
and functions, Malgrange
24.
17,
are given
Chapter VI.
The original
Cartan
where,
proof of
of Grauert - Remmert
27
33.
and K u h l m a n n
21.
see
2 given here
it is a direct g e n e r a l i z a t i o n
in algebraic geometry.
Oka
and
in the
is given
there is an error.
is unpublished,
The proof of T h e o r e m
are given
in Rossi
version of this proof is given I
properties
23.
however,
are due to A b h y a n k a r
sets
sets
in the book by
interesting metric
S. ~ o j a s i e w i c z
i0
2) and in Bruhat -
18).
the t r i a n g u l a b i l i t y of real analytic paper of
sets are
properties of real analytic
and applications Further
C-analytic
15, Example
16,
M a n y v e r y interesting
21;
followed
12.
Chapter V. Cartan
says,
presentation.
Theorem 7 is proved
given
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,
in
A complete
Other proofs 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
to R e m m e r t
IV, P r o p o s i t i o n
formulation
suggested
by
The t h e o r e m H. K n e s e r
20)
been proved
given
28,
Rajwade.
due
to
E.E.
for d o m a i n s
in g e n e r a l
in
in the
be d e d u c e d
(in the case of d o m a i n s
of C h a p t e r
I, P r o p o s i t i o n
H. K n e s e r
20.
of m e r o m o r p h i c C.L.
Siegel
Theorem
36,
37
29;
who u s e d
functions. Grauert
See 3.
generalization functions
38.
see also
important
C n)
at the end,
from an a n a l o g u e functions;
see
on the
were
field
proved by
a very elementary method. 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 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
31.
theorem
It can
The p r o o f g i v e n h e r e
applications
in p a r t i c u l a r A.
in
7).
it has
this reason.
5, and the t h e o r e m
stated
Theorem
seem to h a v e
although
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
functions
IV,
(see also
does not
a p r o o f for
cases of T h e o r e m
deduces
(just as we
22
literature,
We h a v e
proof
o f t e n used.
3 for R e m m e r t ' s
Levi
a more
simple
5 from C h a p t e r
Cn
with
Grauert
15
see
in the p r o o f
form m o s t
image
b e e n used.
Special
included
than
this
30;
in T h e o r e m
M.S.
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 .
Andreotti
-
far-reaching on m o d u l a r of
3;
this
140
REFERENCES
i.
Abhyankar,
S.S.
New York, 2.
3.
Andreotti,
A.
Th~orbmes
de d ~ p e n d a n c e
espaces
complexes
pseudo-concaves,
France,
91(1963),
1-38.
Andreotti,
Bruhat,
A. - Grauert,
H.
Funktionen,
F. - Cartan,
H.
244(1957), Bruhat, d'un
6.
H.
sous-ensemble
Bruhat,
244(1957),
Bunqart,
L. - Rossi.
H.
Cartan,
H.
1961,
39-48.
C.R.
Acad.
Sci. Pari@,
Sur
les c o m p o s a n t e s
irr6ductibles
C.R. Acad.
Sci.
H.
Quelques
propri6t~s Comm. Math.
fondamentales Helv.
H.
On the closure of certain functions,
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