Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, Z(Jrich
83 Oscar Zariski Harvard University, Cambridge, Mass.
An Introduction
to the Theory of Algebraic Surfaces Notes by James Cohn, Harvard University, 1957 - 58
1969 I
9
Springer-Verlag Berlin. Heidelberg. New York
All rights reserved. No part of this book may be translated or reproduced in any form without written permission from Springer Verlag. 9 by Springer-Verlag Berlin" Heidelberg 1969 Library of Congress Catalog Card Number 68-59477 Printed in Germany. Title No. 3689
PREFACE
T h e s e a r e t h e l e c t u r e n o t e s o f a c o u r s e w h i c h I g a v e at H a r v a r d U n i v e r s i t y in 1957-58. As t h e s u p p l y o r the o r i g i n a l m i m e o g r a p h e d c o p i e s of t h e s e n o t e s h a s b e e n e x h a u s t e d s o m e y e a r s a g o , a n d a s t h e r e s e e m e s to b e s o m e e v i d e n c e of a c o n t i n u e d d e m a n d , I r e a d i l y a c c e p t e d a p r o p o s a l by the S p r i n g e r V e r l a g to r e p u b l i s h t h e s e old n o t e s in the c u r r e n t s e r i e s of " L e c t u r e N o t e s i n M a t h e m a t i c s . '.' I r e f r a i n e d f r o m making any changes or revisions, for Ifeelthat these notes can best serve t h e i r p u r p o s e if t h e y a r e p u b l i s h e d i n t h e i r e x a c t o r i g i n a l f o r m . The p u r p o s e of t h e s e n o t e s is t o a c q u a i n t t h e r e a d e r w i t h s o m e b a s i c f a c t s of t h e t h e o r y of a l g e b r a i c v a r i e t i e s , a n d t o do t h a t b y s e l f - c o n t a i n e d , direct a n d I w o u l d a l m o s t s a y - ad h o c m e t h o d s of C o m m u t a t i v e A l g e b r a , w i t h o u t o v e r w h e l m i n g t h e r e a d e r w i t h a m a s s of m a t e r i a l w h i c h h a s a d e g r e e of g e n e r a l i t y o u t of a l l p r o p o r t i o n t o t h e i m m e d i a t e o b j e c t a t h a n d . I s h o u l d a l s o m e n t i o n , i n c i d e n t a l l y , t h a t t h e t i t l e of t h e s e l e c t u r e n o t e s is s o m e w h a t m i s l e a d i n g , f o r o n l y t h r e e of t h e s i x t e e n s e c t i o n s ( n a m e l y , s e c t i o n s 7, 14, a n d 19) d e a l s p e c i f i c a l l y w i t h a l g e b r a i c s u r f a c e s ; t h e r e m a i n i n g t h i r t e e n s e c t i o n s d e a l w i t h v a r i e t i e s of a n y d i m e n s i o n .
Oscar Zariski Harvard University December, 1968
TABLE OF CONTENTS
le
Homogeneous
2.
Coordinate
3. 4.
l~ormal
5. 6.
Linear
7. 8.
Intersection
theory
Differentials
.........................
9.
The
iO.
Trace
11.
The
12.
Normalization
13.
The Hilbert characteristiu function and the arithmetic g e n u s o f a v a r i e t y .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
The
T2
14. 15. 16.
and rings
varieties
Divisorial
of
point
irreducible
an
canonical of
on
a normal
arbitrary
projective
of
on
and
1
.....
3
Theorem
.....
. ..................
II
............
19
. ..............
24
V . . . . . .... . . . . . . . . . .... .... , ........
. ..........
. ..........
systems ...................
.................
lemma..
6
15
. ...........
...............................
fundamental
4
. .........
surfaces
.............
complete
variety
V .........
a variety
genus ...........
polynomials the
variety algebraic
a differentiml
Riemann-Roch
SubadJoint
on
system
arithmetic
....
...... . ...... . .......................
cycles
on
coordinates
varieties.......
~ystems ..... . ..............
Divisors
Proof
non-homogeneous
.....................
29 34 49 52
80 95
Homogeneous an d non-homogeneousP0int coordinates. Let A
n
k
be our ground field; it need not be algebraically closed.
will denote an n-dimensional affine space, and
S
an n-dimensional
n
projective space. These spaces have coordinates in a universal domain. Def. 1.1:
If P = (Xl,...,x)6A , then the local field, k(P), of n n is k(x) = k(Xl,...,Xn) , and dim(P/k) = t.d.(k(x)/k).
P/k
This definition is independent of the choice of the affine coordinates in An/k. Def. 1.2:
If P = (yo,...,yn)~S n, then the local field, field generated over and dim ( P / k )
Note:
by the ratios
YJ/Yi where
is the Yi ~ 0
= t.d.(k(P)~().
T ~ s definition of coordinates in
k
k(P),
k(P)
is independent of the choice of homogeneous
Sn/k.
If we fix Yi ~ O, To see this~ let
then
k(P) = k(yJyi,..., yn/Yi).
Yi = Yo
and assume
Yl ~ O.
Then for any
J,
YJ/Yl = (YJ/Yo)/(Yl/~O)' hence Yj/Yl E k(Yl/Yo""' Yn/Yo )"
Let
Hn_I
(we then say
be a h>~erplane in Hn_1
S n- Hn.1 = A n
is rational over
For a given point Hn_I
Deer. If3:
P e A n.
~
i=O
a Y
i i
P ~ Hu_1.
= O, ai ~ k Then
Clearly then Def. 1.1 and
k(P).
P we may always assume, without loss of generality,
is the ~.3q~erplane Yo = O. If
P a A n, the ideal
I(P)
--(f(X) If(x) = 0 where Def. 1.4:
given by
k), such that
is an affine space and
Def. 1.2 lead to the same field
that
S n,
If P ~ S n,
the ideal
generated by the forms and
dim
in
k[Xl,..., X n]
is the ideal
P = (x)j , and d~m ~ = dim (P/k).
I(P 1 in F(Y)
*--1 + dim (P/k).
k[Yo,....Yn]
such that
is the ideal ~ *
F(y) = O where
P = (y);
-2-
When Let
An
A n = S n - H_l,
we say
An
is an s/fine reDresentative of
be an affine representative of
we can assume
Hn_1
is given by
S n, i.e.
Yo = O.
A n = S n - Hn.1
For any given point
P
Sn.
where in
An,
we have two ideals: I(P) ~
k[XI, ..., Xn]
(P
regarded as a point of
An )
I*(P) C
k[Y o, ..., Y ] (P regarded as a point of Sn). n I(P) = ~F(1, ~ , ..., Xn) I F(Yo' ""J Yn )s I*(P)} .
Clearly Def. 1.5:
The homogeneous coordinates
Yi
of
Pg S n
homogeneous coordinates of
P/k
if
t.d.(k(yL)/k) = 1 + dim (P/k).
Prop. 1.6:
are strictly
If (Yo' "'" Yn ) are homoEeneous coordinates of
PESn, the
following statements are equivalent. (a) The coordinates are strictly homogeneous. (b)
If for a given
i,
0 < i =
transcendental over (c) If
P*
I(~) Proof:
then
is
in
An+ 1 ,
then
is homogeneous.
QJE k(P).
(a) ~
(b), for
Then we can write
are forms of the same degree.
k(y o, ..., yn ) = k(P)(Yi).
aJ = f(y)/g(y)
Hence if we have
9
where
Q)j = fj(y)/f(y)
where
fo' "'" fm;
f and
, ..., cJ ~k(P), 0
can write
Yi
k(P).
denotes the point (Yo' "'" Yn )
It is clear that
Let
Yi { 0 '
f
g we
m
are forms of the same
degree. Assume Assume
I(P~)
is homogeneous, and
COoYim + C~lyi
m-1
Yi ~ 0
+ ... + co = 0 m
preceding paragraph, we can write
~
where
for some
i,
~) ~k(P). J
- f (y)/f(y)
0 < i <=
n,
By the
where the
fj
and
f
J are forms of degree
h,
Hence
fo ( y ) ylm § fl(Y)Yim-1 + ... + f m
and each term has a different degree.
Therefore
J = O, ..., m
J = O, ..., m.
J = O, ..., m,
and so and
Yi
fj(y) = 0
for
is %ranscendental over
fj(Y)Yi m-j ~ I(P~)
k(P).
Hence
aJj _- 0
for for
-3Now assume (b) snd let f(Y) --~=~'~fi(Y) of degree i,
and assume
m Y~ = O. all
2.
be transcendental over k(P).
be a polynomial in
QJm~ k(P),
600 ,
y~
and
Thus
k[Y],
f(y) = O. f(y) = 0
~j = O
for
L6t
where
fi(Y)
is a fcrm
uJ = fi(y)/y i~, m-i+~ , ,
means
O~oy~m+~ + ~ly ~
j = O,...,m~ and so
Let
, then
m+~-I
+ ...+
f.l(y) = O
for
i.
Coordinate rin~s of irreducible varieties. Let V be a variety defined and irreducible over k.
Def. 2.1:
Let
V~
k[X].
An, let
~
If Let
V/k
xi = ~
Def. 9.2:
V/k.
k[X]/~
-residue of
Hence
If V ~ of V
.
Xi,
Clearly
then P6V.
k(V) = k(P).
Sn in
k[Y]/~
k[V],
and ~ k[Y], If
Yi
V/k
We call
is the
P
and
k(V) = k(x).
the canonical ~eneral
is the homogeneous prime ideal
then the homogeneous coordinate rln~ is ~-residue
of
Yi' V/k.
then
P = (Yo' "'"
The function field,
k(v), of V/k is k(P). dim(V/k)--dim(P/k). Yi
k(V),
dim(V/k) = dim(P/k).
is the canonical ~eneral point of
N9t___2e: The
in
k[V].
k[V] = k[x]
We set
= I(V)
of
V
, and the function field,
is the quotient field of
P = (Xl, ..., Xn).
point of
be the prime ideal of
Then the coordinate ring,
residue class ring of
= I(V)
are strictly homogeneous coordinates of
(This follows from Prop. 1.6b.) t.d.(k(y)/k) = 1 + dim(V/k).
P/k.
yn )
-2Prop. 2.3: V C
Sn,
let Hn_ I be defined by
Sn - Hn_ l,
Let V
a and irreducible over Xl, ..., x
Yo = O
and let An =
= V - (V ~ H k,
). Then V is defined n-1 a k(V) = k(Va). The coordinates
and
of the canonical general point of Va/k
can be
n
identified with the Yo' ""' Yn
quotients
Yl , ..., Yn __ , where Ye Yo coordinates of the
are homogeneous
canonical general point of V/k.
3.
Normal varieties.
Def. 3.1: Let V c An.
Let (Xl, ..., xn)
point of V/k of
Q
on V
and let
Q~ V.
is the ring
In other words,
be the canonical general
Then the local ring,
~(V/k),
[f(x)/g(x) If(X), g(X)(k[X],g(Q) ~ O]
~(V/k)
is the ring of all (rational)
functions on V/k which are regular at Q. Clearly k[V] c-Def. 3.2: Let V C
~(V/k). Q Sn, (y) = (y , ..., yn)
the canonical general point
O
of V/k
and
Q~V.
The local ring,
is the ring {F(y)/G(y)IF , G Prop. 3.3: (a) Let V E and let (b)
%(V/k)
Sn,
let V a
Q~ Va.
Then
a k(V),
(c) If V ~ An,
Q~V
and
%(V/k)
of
Q
on V/k
forms of sa~e degree ira kCY],] G(Q) I o. J be an affine representative of V
%(V/k) = C~Q(Va/k). k(V)
and ~
R = k[V], then ~(V/k) -- ~ units in
~Q(V/k),
is the quotient field of
is the prime ideal of .
Q
in
The ideal of non-
is the set ~ R~
functions on V/k which are regular at
of all rational Q
and vanish there.
-5Proof:
(a) follows from Prop. 2.3.
(b) and (c) are obvious.
Remark 3.4: k-isomorphic points of V/k
have the same local ring.
Def. 3.5:
defined and irreducible over k,
If W
is a subvariety of V,
then the local ring, ring Def. 3.6: V/k
~)'w(V/k), of W
4~o(V/k) where is normal at
closed. V/k
Q
Q,
on V/k
is the local
is any general point of W/k.
Q~V,
if
~(V/k)
is integrally
is a normal variety if it is normal at each
point. Prop. 3.7:
If V c An,
Prooft Since V
then V
is normal ~
k[VJ
is integrally closed.
is affine, it is known that
Prop. 3.8" For a projective variety V
k[VS = ~ ~(V/k). Q~ V Q to be normal it is necessary
that each affine representative of V is sufficient that V
be normal, and it
admit~a covering by affine normal
varieties. Proof: This follows from the fact that if V a tative of V, Def. 3.9:
then
If V ~
is an affine represen-
~(V/k) = ~(Va/k), see Prop. 3.3a. Sn,
then V
is arithmetically normal if
k[V]
is
integrally closed. Pr.op. 3.10". Arithmetic normality implies normality. Proof: Let V a ~ V - (V /% H) is given by Yo = O. k(Va)
where
integer
s,
~ YS~o
Then
be an affine representative of V
k[Va] = k[Yl/yo, ..., yn/Yo].
k[Va] , the integral cloalre of
where
Let ~ k ( V )
H =
k[Va]. For a suitable
is an integral function of the Yi'
Hence YoS ~=
F(y)~ k[yJ. Now
~ = f(y)/g(y)
y~F(y) - g(y)F(y) = O, F(y)
where
f
and
g are forms ~f degree
t.
Since
each homogeneous part is zero. Thus we can assume
is a form of degree
s.
Therefore we can divide by
yS and so ~E k[Va]. o
6-
4.
Divisorial c~cleson a no!mal prgJective variet~ V/k (dimlY) = r ~ I)
Defo 4:1: A ~rime diviroria! cycle on V ~ such that W
Def. ~.2:
is a subvariety W
is defined and irreducible over
and
1.
~(w)
= r -
If K
is a finitely generated extension field of k
t.d.(K~) = r, then a prim~ d$visor of K/k v of K/k
If ~-
r - I
has tran-
over k.
is a prime divisorial cycle on V,
is the valuation ring of a prime divisor Furthermore,
and
is a valuation
such that the residue field of v
scendence degree Prop. 2.3:
k
of V
then
(V~) F k(V)/k.
of F is a discrete, rank I valuation; and
v
v
~
C the residue field of v
is
k(~-).
r Proof: Let H V
be defined by Yo = O.
= V - (V~H)
and ~ a
d i m ( ~ a) = r - I. noetherian.
If ~
ra
because
R
is.
" ( [- ~
Let R = k[VaS; then R is the prime ideal of
minimal prime ideal in R
(v
= ~
Since we can assume ~- ~ H, let H).
Then dim(V a) = r and
is ~utegrally closed and q
in R, then
since
is a
r
where
is integrally closed and noetherian
Furthermore, since ~
only proper prime ideal in R~
~
is minimal in R,
~R~
. We have thus shown that R#
Dedekind domain with only one proper prime ideal.
Hence
R~
is the is a is a
discrete valuation ring of rank I.
Since
- k i t a]
we have the last statement.
is the quotient field ef R / ~ ,
-7The following is an easy consequence of Prop. 4.3 (but we shall not prove it here): Cor. 4.4: With the assumptions and notations of Def. 4.2, every prime K/k is a discrete, rank I valuation, and its
divisor of
residue field is also finitely generated over Def.
If
4,5:
~ }/ O,
~k(V),
on F- . [-
the integer
(~)
v [-
k.
is the crder of ~___ '
is a prime null cycle of ~
if
v
---
a prime polar cycle if
v
_
(~) > O,
F-
If ~
(~) < O.
~ O, we define
V v
r
(o)
prop. 4.6: Let
+-
Q~V c
of ~ Proof: Let ~ = prime ideal ~
.
Sn,
and let ~ e
k(V).
passes through Q, then ~ ~(V/k). Q
~.
of
is regular at Q.
A prime cycle ~By assumption
If no prime polar cycle
corresponds to a minimal
~ ~r
hence
z . ~ t~2ough the minimal prime ideals of
~.
is integral~closed~ and so ~
. Hence
Frcpo 4,7:
If ~
~
= ~
Since V
has no polar cycles, then ~
/ ~
~
where
is normalj
~ ~ ~o
is a constant, i.e.
is algebraic over k. Proof I:
If ~
any valuation of over
k,
has no polar cycles, then ~ ~ ( V / k ) Q~V > k(V)/k, then v(~ ) = O. Assume ~
and consider ~
where
there exists a valuation v the principal ideal contradiction.
Hence ~
Proof 2: We have given by Yi = O,
kKv]
o
(~).
/~
Q~Va
~ = i/~
v(~) > O,
is algebraic over % ( V a / k ) o kEVa].
and
k[V(a~
and M v
contains
v(~ ) < 0 which is a
Let
= k[xl, ~
Hi be the hyperplane where x n]
is
is transcendental
k.
and let glijt, ~ g -~/~Hi) a
o' "'" Yn ]
v
. By the extension theorem
such that R v ~ k[ ~] Since
I and if
i = O, , . . , n. where
x i =yi/y a.
-8By Prop. 4~
and Prop. 3.3, we have ~ ~ k[Xl, ..,, xn],
= Fo(y)/yS i
where
we can write
in the y's.
F~
is a form of degree
~ = Fi(y)/y i.
Let
Hence
s.
~ Yi
m = (s - l)(n + !) + i,
and let Then
as a form of degree m
= ~ c
gives an equation for ~ Prop. 4,8:
If
over
~ 6 k(V),
~
~
k, { O,
Similarly, for each
is a form of degree
be the monomials in the y's of degree m. in the y~st
namely
I
then ~
hence
co1 , ..., cJN
~ ~
can be written
~%,
c~E
k.
- 6 ~2~ ~I
o
s
This
,=o.
has only a finite number of
prime null cycles (or of prime polar cycles ). Proof: V
Use the fact that any hypersurface, not containing
V,
cuts
in a variety which has di,~ensien r - I, and which therefore has only
a finite number of irreducible (r - l)-dimensional components. Def. 4.9:
A d i~:tsoria! cycle on V/~.
is an ~lement of the free
(aadx~zve) group genera5ed by the prime divisorial cycles. If
Z = 7mi~- i,
the sum being f~_nite, end if
is a ~rime component of
Z.
Notation:
mllo,
[Z] =
Fi
mhmiJ0F i "
>
If all
mi = 0
effective if D~f. 4.]D:
If
and some Z > 0
~ s k(V),
([) = Z r
~
or
~ ~ 0,
w (~)F r
v (~)F (~)>o r
.
m i > O,
write
Z > 0;
Z
is
Z = O. is
then the ~:_d~wisorial..cTcle, o f The null divisor of ~
. The ~ o l ~ ~ v i ~ r -
of ~
is
-
F
-z Note:
(~ ) = o ~ If ~
v ~
r
(~)F
.
is a const~,~.
is not a constant, then neither
effective.
(~)
= ( ~ ) + (~).
(~)
nor - ( ~ )
is
-9Def. 4.11:
Two divisorial cycles
ZI
and
Z2
are linearly equivalent
w
(notation~
Z1 = Z2) , if
Z1 - Z 2
is the divisorial cycle
of a function. 7~ro~, ~.12~
Given
Z,
the set of all functions
~ E k(V)
such that
>
(~) + Z = 0
is a vector space over
We denote this space by Proof:
~(Z).
This follows from the following two facts~
Prop. 4.13|
If
Z I ~ Z2~ * ~,
onto Theorem h.l~
and if, say,
~6~(ZI)
,
The space i(Z)
~(Z)
such that ~
Clearly
Z
ZI - Z 2 = (~),
then the map
is a linear isomorphism of ~(ZI)
We can assume
~(Z)
Then we can write
Hence
X m Z
~ (0). Fix (~) = X - Z
and so ~ ( X )
=
~(Z).
where Thus
is effective.
k ~ ~ ~
k(V)
c g k,
is finite dimensional.
~ 0.
X ~ 0, i.e. X is effective. we can assume
for
2).
Proof (for the case r = 2 only):
because
k.
is.
k(V),
Since
hence
~(0)
~
= ~,
is finitely generated over we can therefore assume
k
Z ~ 0.
Then we can write h
i=l
>
mtC i
' mi
7o
We shall prove the theorem by induction on We can assume that there exists since otherwise
~(Z) = ~ ( Z t)
where
f
in
'
h l.
Z mi . ~(Z)
such that
ZV = ( m l - l ) q
and the result foll~s from the induction hypothesis.
v (f) = -mlJ r I + m 2 F 2 +...+ m h r h
-lO-
Let
M = [ ~/f I ]~g~'O(Z)]. Clear~y
denote the elements of M
by ~
dim(M) = dim(~(Z)).
. Since
v
(~) > - m I,
C
we have
Vl_l (~) ~ 0 SLnce Vl_l(f) = -~l" Let Trc~(1 ) = ~k(r-1) the restriction of
71
-*RF
/M
to [-I ' i.e. the where
vr
Let us
denote
-r~sidue of 9? under the
1 is the valuation ring
R
q
w
{zs
Iv (z) > O ~ and M F r I I
is a .mapping of M
onto a s/bspace
{
z,
(5-1)F
=
induction hypothesis,
ker(~)
sufficient to prove that M Let ~-l
(f) + Z = Y.
2
k(~'l)/k
where
h
+ .,. + ~ U
is finite-dimensional.
ker(~) =
. By our
Therefore, it is
is finite-dimensional.
Clearly Y > O.
set of (algebraic)points.
;(Pi )
ef
+ m2 ~
is not a prime component of Y.
G(y)6 k[y](= k[V])
M
Since
v (f) = -ml, we see that F 1 Hence ~ l / ~ [Y] = { Pi }' a finite
We can construct two forms
F(y),
such that
all i; F / 0 en ~ I
o
; F # 0
on each prime
component of Y (2) Since
O = 0 ~r
on each prime component of and
G~
we can assume
F
values for
and
r
component of Y each
P..
Y;
G /0
on 5" 9 1
(r, s positive integers) still satisfy (1) and (2),
and
G
have the same degree by choosing appropriate
s. We have
G/F
is a n,~ll cycle of
in
k(V).
Furtherm~e, each prime
G/F. Finally,
G/F
These properties also hold for any power of
is regular at G/F. Hence
i
there exists an integer
s
has the same properties as
which is large enough so that G/F
and the null cycle of
~
~ = (G/F)s is
=> Y.
-3_1-
We can write Then
~ Let
(~) = Y + X - X' where X, K' > O. Let ~ O, - and
~k(~
~I' '''' ~ q
I).
(~).
rl
be the prime divisors of k ( q ) / k
project into the set {Pi]~
(ii) ~(@)
= Tr
We shall show: if ~
~ o if ~ ~ ~i' i-l,
M,
whiQh
then
..., q
and this will prove the theorem (in view of kncwn facts in the theory of algebraic functions of one variab~ ). me
cente~ of
~l
is m
the set I P i ~ "
( ~ ) - ~ + X - X', we see t ~ t
Fr~
(~
(V~) ~ ~' § X - X'.
- Y' - ~,
Smoe
P1 ~ K ' J '
-
~ is regular at Pl' ~
and v~
)=>
is also regular at Pl~ Hence v ~ l (
) > O~
(). ~
Let Q be the center Then Q~[Y] is regular at
of ~ on q
because Q ~ P i } Q.
Henae ~
where
and Q ( q .
~ ~i' i = i, ~176 q.
Since
is also regular at
Q,
( ~ ) = Y ! - Yj Y! > 0,, and v & ( ~ ) > Oo
5. Linear~s~stems L~t to Z.
~Z~ denote the set of all effective cycles linearly equivalent If Z t z Z, then clearly
then we can associate with #
IZt~ = ~Zi . Let ~ 6 ~ ( Z ) ,
the effective cycle
(~) + ZaIZl . It
is clear that this map ~ : ~ -~ (~) + Z maps ~(Z) k
~ ~ O,
onto ~ZJ. Assume
is maximally algebraic in k(V). Then ~ ~ = ~i~ if, and only if, = C~l where
c C k (since c = ~ vj.
into a projective space over k.
6k(V)).
Hence we can turn IZl
-12-
Def. 5.11 The set I zl
together with the projective structure defined
by ~ : ~ * (~) § z, ~
~(z),
~ ~ o, is called a
complete linear s~,stem (the oemplete llnear system determined by the cycle Note;
Z).
The 1-dlmenslonal subspaces of ~Z I are called ~ . Viewing
~Z~
as a projective space, we have dim I Zl = dim ~ (Z) -I,
and dim ~Z| = - I ~ Izl Let
(whence I z'l -IZl ), let
Z' = Z
~' ~ ( Z ' ) , and define
is empty.
$' #O, and let ~ = ~'.
~' :6 ' * ( ~ ' )
(~) = Z - Z', ~ # 0 .
Then ~ g ~(Z)
~mp!ies
~g
Let
~:
~(Z').
that the projective structure is independent of the choice of Note ~lso that
Z ~ Z'
" z', ~ -~
,
This shows Z
in ~ Z|.
i~r.plies | Z lf] ~ Z'I = ~.
Def. 5.2: A linear system is subspace of a complete linear system. Prop. 5.3:
If L
is a non-empty linear system and X
is any cycle
which is linearly equivalent to the cycles of I XI, then the set ~(L,X) = { ~ i ( ~ )
L~
+ X ~ L or ~ = 0
is a finite dimGnsional subspace of k(V)/k. k c ~ (L,X) ~ X~ L. subspace
M
(~) + X ~ 0
Prcof: Since
L
~(L,X)
such that
then the set
system L and ~ ~ ~(L,X). is a subspace of
is a subspace of
is a defining function ~odule of
Z = Zr m F ( Z ) F
X
~(X)
The second assertion is obvious, and
the last follows from the fact that M
Notation:~ Let
~ # O,
~ i o~ is a n n e ~
is a subspace of X,
~(L,X)
and given any cycle
for all ~SM,
which proves the first statement.
We say
Furthermore,
Conversely, given a1~y finite-dimensional
of k(V)/k,
~(~) § xl ~ ,
L, i.e.
~ (X). L.
) m r(L) = min~mr(Z)~ Z6L).
-13Let
IXl. Let ZI and Z 2
L be a linear system contained in
be linearly independent cycles in dependent on
and z2.
ZI
and
~en, if
we must have
Z2,
L}
i.e. Z
and let
Z~ L be linearly
is in the pencil determined by
ZI
(~l) + X ~ Zl, (f2) § X - Z2, ~l' ~2 ~ ~(X),
(el ~i + c2 ~2 ) + X = Z
cycle ~- , we have
mr.(ZI) = v
for some
(~l) + m
r
Ol, O2 mk.
(X), m
r
r
For any prime
(Z)
= v (~2) + m (X), r r
2
and mr.(Z)= vF.(cl~ I + 02 ~2 ) + mr(X ) .> rr~n~.Vr( ~i), Vr( ~2)~ + mr(X). Hence m r (z) .~ rain{ Assume
X "- X'
mr(ZI), mr.(Z2 )~. while
X ~ X'.
Since m (X)= mr.(L)- min{vr(~)I~L;O], r
we see that .~ (T,X) # ~ (T.,X,) Fo~ded Prop. 5-~: Let
~ r ~.
L be a linear system, and let To
Z + Y o >" 0
for all
Z
in
L.
be a cycle such that
Then the set
L' = { Z + Y J Z ~ L }
is a linear system and dim L = dim L'. Proof: Fix
X
such that
L C IX I.
Let M = ~(L,X).
Then
L' =~(~)+X +YoI~6M, ~#0#, and M-I(L,,X+Yo). Def. 5.5: A positive cycle L
Cor. 5.6:
If
if
Z > .B
B
is a fixed con~onen%, of a linear system
~ " all ~o~
Z~ L 9
B is a fixed component of
L,
then the set [ Z - BIZEL3
is a linear system having the same dimension as lh-oof: By definition
Z - B ~ O
for all
Z (L.
L.
The result no~ follows
fk'cm Prop. 5.i~ ~,ith -B = Yo' i
Le~. M
be a finite dJmansional subspace of the vectcr space k(V)/k.
Then there exists a smallest cycle ~ M.
Let
Xo
such that
Lo = ~(~ ) + X o ] ~ M , ~ # 0 ] .
component, and we have M = ~(Lo,Xo). whi=h admits M
If L
Then
(~) + X o > 0
for all
L o has no fixed
is any other linear system
as defining function module, i.e., if M = ~(L,X),
X = X o + B, B > O,
and B
is a fixed component of
L.
Thus
Lo
then
is the only linear system, _free from fixed components, which admits as defining function module.
We will denote
Lo by
Theorem 5.7: (Theorem of the Residue (Restsatz)). system and Yo
Proof: Let
L:
is a linear system.
L I = [Z[Z - Y o "> O, Z~L}. ~
Then
~C
Z belong to the pencil determined by
We have
vp(Z) _~ mJ_n~v (ZI), vr(z2)), hence
Zl " Yo
and
Z2 - Yo
are effective,
L
Let
Hence
X~L' , ~howing that
X ~ Xo, X ~ O.
ZI
and
Zl, Z2~LI,
Z~~ in
~ZI~ .
T.hus Z~Ll~and this shows
Then X o = Z - Y o
L
instance, take = Tr ~ C . SinGe
L.
Fix a cycle
and such that ~Zo
in
L.)
~-
arid so X + Yo ~ L.
Zo
a prime cycle which linearly equivalent to
is not a component of
Let M = ~(L,Zo) ~- [ ~ [
for ,an5- ~s M ( V
where
L' = IX o~ .
L be a non-empty linear system, and
the cycles in
such %hat
is complete,
L and, by Prop. 5.4,
Then X + Yo -- Xo + Yc = Z
is not a fixed ccmponent of
the ~
L
Z - Yo is effective since
is complete, and let Xo~ L'.
Z~L.
Let ~
If
is a linear system.
Assume
Let
L be a linear
is a linear system. Let
and let
LI
Let
~
it is sufficient to prove that
that
LS(M).
an effective cycle. Let L' = { Z - Y o l Z ~ L '
Z -> Yo) " Then
the
M
# |
arA lzt M
Z o.
(For
(~) + Zo~ LJ. be ~ e set of all
is ~ot a fixed co~poneRt of L, there exists
(~) = X - Zo
satisfies the condition
Tw
~ ~ O.
X~ L
Hence
r LS(M)
is a linear syshem on
easy to s,~e that s.hall refer to Zr
!
L~(M) LS(M)
~- without fixed compo~,nts.
depends only on
L
and
as the rec~J.ced trace of
f-- and not on L
on
It is
Zo~ We
(notation:
-15Theorem 5.8: Let
L' ={ Z - C ~
are as above.
Z.~L, Z
Let
C
@. 0 3 where
L = Red Tr
L.
L, C
Then
r dim L = dim L I + dim ~ + 1.
6.
Divisors on an arbitrary variety Let
Let
v
V~
A
i.
Let
V.
variety with v
zi
k(V)/k.
and
Def. 6.1:
Let P
Then for some
be the v-~esidue of
be the residue field of
belongs to
on
(Yo' '"' Yn ) be a general point of V/k.
be a valuation of
for all let
Sn, and let
V.
W/k
v.
W
yi/Yj, i = O, ..., n~
Clearly the point
be the locus of
as general point over
is called the center of
P
over
k (W ~
and
P = (Zo, ..., z n)
k, i.e. the irreducible
V).
Then
W
depends only
v
on
V.
The following properties characterize the center on
~(V/k)
~ Rv
where
(b) w(V/k) Let
W Let
be the center of ~
= %(V/k),
is equal to
Rv
v
is the valuation ring of
v
on
and let
~e have
V.
~e know that
F(y)/G(y)~ ~ , G(z) ~ O,
v.
F(z)/G(z) { O,
(~w(V/k) =
where
F
and
G
and the v-residue of
showing that
Parthermcre, this v-residue is zero if and only if and only if
of
(v/k)'
are forms of like degree s. F(y)/G(y)
W
V:
(a)
%(V/k).
v(Yi/Yj~ .~ 0
V.
Prop. 6.2:
Proofs
J,
F(y)/G(y) s R v.
F(z) = O, i.e., if
F(y)/G(y) ~ ~
. Hence conditions (a) and (b) are satisfied. W Conversely, assume that conditions (a) and (b) are satisfied. Let
(Zo,Zl, ..., zn)
be a general point of W/k.
of generality, that
Yo ~ O.
Then
We may assume, without loss
yi/Yo 6 ~W (V/k) c
R~,
showing that
-16 -
if
z'.
is the v-residue of
yi/y ~
and
~'
is the center of
v
on
1 ,
V, then
,
!
(Zo, Zl, ..., Zn)
is a general point of
is a homogeneous relation, of d e ~ e e ficients in
k,
then
s,
f(y)/yS E~T~ C
Conversely, if
f(z') = O,
f(z) = O.
This shows that
Def. 6.3:
Let
Va
,
whence
If
v
V,
then
Va
If
zls,
f(z) = 0 with coef-
f(z') = O.
V
then
f(y)/y~ E M v
~Yw(V/k) = ~ W '
whence
~ = W'.
be an affine representative of
finite on Prop. 6,4:
between the
M
O
~'/k.
if
V.
Then
v
is
k[V a] C R v.
is finite on W ~ Va
Va,
and if
~0T is the center of
is an affine representative
called the center of
in k[va] is
v
~
V a.
Wa
v
of
The prime ideal of
cn
W
Wa
k[Va]. V
Proof~
The first statement is obvious, and the second follows from
Prop. 3.3 and Prop. 6.2. Since v
k c
A
and
is a prime divisor of
Prop. 6.5:
If W
d i m V = r, k(V)
we have
is the center of! v
on
V, then A.
k(W)
Hence
dim
can be W ~ dim v.
This follows directly from Prop. 6.2.
Def. 6.6: If
- I;
if t.d. A/k = r - i.
identified with a subfield of Proof"
0 .< i.d. A/k ~ r
v V/k
is of the first kind with respect to
V
if
dim ~'! = dim v.
is a normal variety, then we have a (l-l)-correspondence f
between the (r-l)-dimensional irreducible subvarieties
~-
of
V,
i.e.,
!
the ~rime divisorial cycles, and the prime divisors
v
of the first
kind with respect to
v
if and only if
~
corresponds to
If
V
is not normal, we shall continue
to call "prime divisorial cycle" of
V
any (r-1)-dlmensional irreducible
is the center of
V, v
where on
V.
-17-
subvariety of V/k, of
but this time we may have several prime divisors
k(V)/k which have a given prime divisorial cycle of V/k
Theorem 6~7:
(a) Let
~-
as center.
be a prime divisorial cycle on V/~.
every valuation of divisor of
k(V)
with center
V
Then
is a prime
k(V)/k.
(b) There exists at least one and at most a finite number of prime divisors of on
V/k.
k(V)/k
having center
Furthermore, there is only a finite
number of F ' s
on V/k
which are centers of more
than one prime divisor. (c) If
V
is the center of a prime divisor
: k( F ) ] Proof:
|
9
(a) By Prop. 6.5 we have dim F
we have dim v = r-l. Therefore
v
v, then
~ dim v.
Since
dim ~
is a prime divisor.
(b) The existence of one prime divisor with center known result.
Let
~
= r-l,
be the integral closure of
~=
~
is a well-
~(V/k))
in
i
its quotient field. is a finite
Let ~
be the maximal ideal of ~
dimensicnal vector space over the field ~/~
~/~is
~'-module, the residue class ring ~/~
.
a finite
Therefore the ideals of Hence
satisfy the descending chain condition.
9 Since
~/~
has only
a finite number of maximal ideals3 and since each maximal ideal of contains ~ ~, we have shown that ideals.
If
v
~
has only a finite number of maximml
is a prime divisor with center
F
and
maximal ideal, then M v / 7 ~ is a maximal ideal of ~ , contraction in ideal of
~
~r is the maximal ideal of ~ . , then the quotient ring
~
If ~
Mv
is its
since its is a maximal
is a discrete valuation
-18ring, and the corresponding valuation of and is the only prime divisor
v
k(V)/k
such that
is a prime divisor,
Mv
=
This
proves that there are only a finite number of ~-Time divisors with center
~- .
If H
is a hyperplane in
Sn, V ~
H,
then
a finite number of prime divisorlal cycles.
H D~
contains only
It is sufficient, therefore,
to prove the last part of (b) for prim~ divlsorial cycles of the affine representative
Va = V -
(V /T H) of V.
be the inte~'al closure of
R
R
~ ~I' "'" ft , g~R.
which contain
g,
does not contain
~-l' 9
and let
If
K"
, ~-s
be the prime diviscrial cycles
is any prime divisorlal cycle on V
[-i' then the mini~l prime ideal ~ g.
R'
Let ~ l' "'" ~ s be the minimml prime ideals of
determined by ~l' ""' ~s" distinct from the
R = k[Va] ~ and let
in its quotient field. Since R' is a t R' = ~i=l R ~ where ~ = fi/g, i = l, ..., t,
finite R-module, we can write with
Let
Therefore
~
(V/k) D R',
at ence that if we set ~' = ~ F ~ R',
then
of
~-
in
R
and from this folluws ~w(V/k) = R'~,.
Hence
g (V/k) is integrally closed in its quotient field (which is the r" quotient field of R). Hence ~- is the center of exactly one prime divisor.
This p1'oves (b).
(c) is obvious. We wish now to extend the results of ~4 to arbitrary varieties
V.
However, we have already observed that a prime divisorial cycle may be the center of more than one prime divisor (of the first kind with respeot to V); hence the free group generated by the prime divisors is no longer isomorphic to that generated by the prime divisorial cycles. Therefore, we must now replace the "prime divisorial cycles" of w
by the "prime
-19divisors" (of the first kind with respect to V), and the elements ("divisarial cycles") of the free group generated by the former, by the elements ("divisors") of the free group genera bed by the latter. For clarity we restate Def. 4.5" Def. 4.5':
If
~s
of
~
~ # O, on
v;
v
the integer
=
is the order
>o,
is a prime null divisor of
a prime polar divisor if
v(O)
v(~)
v(~) < O.
If
~
=
O,
we
define
+-.
With the above indicated change, all the proposi%ions of ~4 remain valid, except Prop. 4.6, ~ which the conclusion must be replaced by
7.
"~
" ~
is regular at
belongs to the integral closure of
Q':
~ (V/.~)".
Q
Yn.tersect~;on,.theery on algebraic surfacec (k algebraically closed) Let W ~ V
where
dimV = r
and
d i m W = s.
It is well known
that we can find maximal chains W < W I < W 2 < ... < V varieties
Wi
where
dim W i = s+i. Hence, if
the Krull dimension of
D~f. 7.!:
W
~
~=
of il~educible ~
W
(V/k), then
is r-s.
is a s~mple subvarie~ of
loc~l ring, i.eo, if ~
v/k if
= A~w(V/k)
is
has a basis of r-s elements.
Equivalently
Let
P~V
and let W
be the variety for which
general poi.nt. Then we say that Cot. 7.2:
If V
is normal and
P
dim W = r-l,
P
is simple if W then W
is is.
is a simple sub-
varlsty of v/k. S~ce a regular local ring is integrally closed, we have Car. 7.3: Any n~-singular variety is normal.
-20If
Z = Z m (Z)~- is a divisorial cycle on V, and if F P~V, then the local P-component Zp of Z is ~ P a W m r - (Z) F
Def. 7.4:
We say
Z
is locally linearlY equivalent to zero at
Zp If in
•pe
if
ction
exists, it is determined to within a factor which is a "omit We say
Let V through
for
P
" ~ = 0
is a local equation of Z at
be non-singular, let
P.
Then
I~
= 8rpt, i.e., ~
= ~
P~V
P."
and let ~- be a prime cycle
is a minimal prime ideal in
is principal, mud
(t)p = F
9
~p.
Hence
(we recall that the
local ring of a simple point is a unique factorization domain, hence every minimal prime ideal in that r~ng is principal. ) Prop. 7.5" On a non-singular variety V,
every divisorial cycle
linearly equivalent to zero at every point Proof:
Since
Zp = ~
m
P~ F the preceding paragraph.
g
(Z) r- ,
Z
is
P.
the result follows directly from
Let F/k be a non-singular surface. Def. 7.6: Let ~-,A be distinct irreducible curves on F/k. and let ly at
~ = O, ~ = 0 be local equations of P (~, ~ d ~ p ) O
and we call the integer
If
P
F
~-,A
P EF,
respective-
Then we define ( ~=
i(~-,A;P) = di~( ~/0~( ~ , ~ ),
multiplicity of
Let
i ( r ,A;P)
and A
is not a common point of
the intersection
at the point ~-
~), P
and A,
P.
then either ~
or
is a unit in
d~ and the intersection multiplicity is zero. If P is P a common point of F and A, then P is a rational point, dim P/k = 0, dim ~ =
2
and
~(~,V
) is an ~-primary ideal ( ~
C
~(~,~)
-21-
Cot. 7.7: i(F-,~;P) >- 1 ~ P~- F-/) A. Let X = ~i 9
ai~i' Y = ES b~A~ be divisorial cycles without J=lO co~on components. Then we define i(X,YIP) = __Ei,jaibji(~i'AJ;P)" We define the intersection number~.Y) = __ % s Fi(X,Y;P), hand side is a finite s~o since X
and Y
where the right-
do not have any common
components. Theorem 7.8:
P
Let
be a prime cycle, let
~ O,
such that
~-
= Trr ~ ( ~
Let ~
degree of
P~F
and let
~k(F),
is not a prime component of
s
~ # 0).
(~) . ~ere, if
Then
Vl, ..., v
of k(F')/k which have center
P
(~).
i((~),ViP) =
are the valuations
on
and if
vi(
) = ni,
then deg (~)p = nl+n2 + ... + n .S g
Proof: We can assume
~6~"
local equation of [-
we have
i((~ ) , P I P )
~p([-/k).
Let
at :
~'
~ = %,
Let
P; then ~
and let
~i
= R
vl
for
~o/o'(~,~)
= dimkll~ ~
be the integral closure of ~
a D e dekind domain.
i = I, ..., g. Therefore
Each
~I
in
where
in
=
... g
k(~'); ~I' """' ~ g ~'
where is
. A well-known
= dimk ~ ' l ; ' ~
dimk ~ ' / ~ < |
~.
:
~ i s maximal, and so
result allows us to conclude that n I + ...+n Me shall show in a moment that
= 0 be a
is the prime ideal of
a semi-local ring with a finite number of prime ideals
~'
~
: deg (~)p.
Assuming this, we
have
dimk~'/~: d i ~ ' / @ ' ~
+ d h % ~ ' ~ / ~ : ~mk~'/~' ~ + d~k~'/S- .
Furthermore dimk B ' / ~ ~ = dimk ~ ' / & :
+ d~
~/~ ~.
Hence
-22-
Thus
i((~),p--;P) : d ~ ' l ~
:
We ~rast now show that d ~ such that ~ '
= ~
such that
~ ~ . Let
c~
~ o D i.
It is easy to see that
~
d~a'/a'~
~' I ~ < =.
in 7'le
).
Since
c E~,
~
c
be the set of all ~ach
is an ideal in both ~ '
dimk~'/~
+ ...
+ rig.
There exist (~I' "''' ~Oh ~ 6'
Hence there exist elements
is the largest set with this property.
of
-- deg (~)p = n I
( J~ < =,
c #0~
including zero.
and ~
, and that
is called the conductor we have, a fortior~,
< =
Cot. 7.9: If YI m Y2
and if Yi
i = I, 2, then Proofs ~Te can assume X
has no common component with X,
(Y1.X)= (Y2.X). is a prime cycle
~- and extend the result by
linearity.
Let Y = YI " Y2 = (W)' and let
deg (~)p,
and
P~-.
Then
i(Y,~-~P) =
(Y.F)= ~
i(Y,~,P) = deg (~). ~at deg (~) is P( F simply the sum of the orders of the zeros and poles of ~ , which is zero.
Def. 7.10." If Y
and X
are any two cycles, then we define the inter-
section number
(Y.X) to be (YI.X) where
YI ~ Y
and Yl,X
have no cemon components. By the preceding corollary this defi~zs of YI"
The existence of a Y1
ms independent of the choice
follows from the fact there there exist
f k(V), ~ ~ 0, of X.
such that v (~) = - m (Y) for all prime components r F If we set Y1 = Y + (~)' then Y1 satisfies the requirements of
Def. 7.10. We call (X.X) = (X2)
(x+Y.X§ a
§ 2(x.Y) §
prime cycle and
the self-intersection number.
(y2) and (~)2 ,,n2(X2). If
dim IX~
>- I,
then
(X2)
Clearly X
-> 0 because
is
-23-
dim ~XI >- I
implies there exists
Y -> O,
such that Y =- X
and Y # X . In addition to the divisorial ~cles considered up to now, we shall also deal now with zero-dimensional cycles (brieflyl zero-cycles): these are the elements of the free (additive) group generated by the points of Def. 7.111 Let X
and
Y
be cycles without common components.
the .intersection .c~cle of
X
and
Y
(denoted by
We define
X.Y)
to be
Z i(X,YIP)P. Let
Z = Zpm~
be a zero-cycle.
Defining
deg Z = Y m p,
we get
(X.Y) = deg X.Y. Let
~
be a prime cycle and
component of X. PEp. 7
Then
(~)p
~
Let
~P = O
Trv~P = ~ Thus
X.~
X
a cycle such that ~
be a local equation of k(~).
We define
X
Trr.X by
is a O-dimensional cycle on
F,
is not a at
P
where
(TrJ)p= ~.nd Tr X
P is a O-dimensional cycle on Let
(2)
if
.
L be a linear system on
component of (I) ~
~
L.
Let ~ = CTr
is a linear system on B
F
X~EL,
is not a fixed
and assume
[-- not a component of
L~
then
L -
B
= Red
Tr
L
r m
We define
Tr L to be
Cor. 7.12: dim L = dim Proofl
Then
X~.
~- ,
is the fixed component of L .
Tr L + dim ( L - ~-) + I. P r Obvious (Theorem 5.8).
F.
.
88.
Differentials Let
k
K/k be an algebraic function field of r
variables ~here
is algebraically closed. We consider derivations of
maps
D
of
K
into
K
satisfying
D(x ) D(xyj Dc for all x,y
in
derivations of definition,
K
K/k, i.eo,
Dx + Dy, =xDy = 0
and all
+yDx, c
in
k.
Let ~
denote the set of all
K/k~ it is a vector space over
(~D)(x) = ~(Dx)
for all
K
in virtue of the
~ and x
in
K.
We state (without proofs) two well-known facts: (I) ~ (2)
is an r-dimensional vector space over
K.
If ~ Xl, ..., XrB
then there exist r
is a separating transcendence basis of K/k, xj derivations P/~ x i such that ~ = 8ij (i, J ~ l,
..., r), and these derivations form a basis of ~ if the derivations
over
exist, then ~ Xl, ..., xr~ xi transcendence basis of K/k. Let
~
~(Y* denote the dual space of
Def. 8.1, A differential of
K~
@~
q
If
tEK,
dr(D) = D r
we define
for all D E . ~
dr, ~
is a separating
is an element of ~ * .
is a multilinear antisymmetric
function on the direct product, times with values in
Conversely,
.
of degree 1
A differential of degree
K.
~q,
of @ ~
with itself q
K. the differential of If
tl, ..., tqEK,
t,
by the equation
we define
dtldt2 ... d %
by (dtl... dtq)(D1, ..., D q ) = det (Di(tj)). If ~Xl, ..., x r~ any differential
is a separating transcendence basis of CJq, of degree
q,
can be written as
~q = il~i2 Z~ ..~iq~ "I' "~ ,iqdXil...dXiq, ~l,...,lqr K.
K/k, then
-25For any differential d~
of degree
q Di,...,Dq+ I)).
exists a differential P
~q-1
such that
V/k.
Let
tl, ..., tr
(i.e.
ve define the differentia]
d ~q-1 = ~q"
be a rational simple point of an r-dimensional
variety let
q
by
a~q
Let
of degree
q§ (-l)i(~q d~J (DI, ..., Dq+ I) = Y ~ (DI, ., q i=l i "" is closed if d ~ q = O; CJq is exact__if there
q+l
A
Theorem 8.2:
Oq
~ = ~p(V/~), ~
= a~p(V/k)
and
be uniformizing parameters of
m~ =
Y ~ ti). i=l transcendence basis of
P
on V
Then ~ tl, ... trl is a separating K/k,
and
a. ~ C ti
~
for
i =
I, .., r. Proof:
To prove this theorem we shall use the following well-known result
(which we state without proof): (3)
If an extension
over
k ,
k(x)
of a field
k
it is separably algebraic over
In view of (3) we must show that if tr)
(hence
D
is trivial on
is the quotient field of on
~
has no non-trivial derivation
.
If
fi(tl, ~
~ ~ tr)
, in
~
and conversely.
D
is a derivation of
k(tl, ..., tr)),
then
D = O.
K/k(tl,... , Since
such that
i > O,
A I O,
assume that trivial on Therefore Let A I O, and
such that D ~ c
~ - f i ~ , i+l,
= O
~.
It follows that
k(tl, ..., tr) , D~
D
Q'.
(AD)(~) C
because
be any of the
such that if we set
we have
D(~
Since
~
Replace
D
by
D~+lc~i.
AD.
is the
- fi ) = D ~ g i
then
D~t. _ O (mod A) in ~ . This implies J Hence D ' ~ a / ~ ~ ( A + ~ , i) = ~A. Therefore i=l
D I~
~
~ .
A
D
is
for all
Then
A
i.
in D ' ~ l"+ l C ~ :
D'k[t] C A k [ t ] C A ~ D~E
of
So we may
Since
/~ ~qi = (0). i=l S / S t i. There exists an element D' = AD,
D
there exists a polynomial
quotient ring of a finite integral domain, there exists an element ,
K
, we need only consider the effect of
then, for any
kit]
k,
.
i
-26-
Def. 8.3: A derivation DRv C R
D
where
v
is regular at a given valuation R
is the valuation ring of
V
Def. 8.4:
K/k
v
form a
if
R v. A differential
~Jq
for all DI, ..., D CO q
is regular at v
v
D v = ~D ~ I
of D
if
~q(Dl, .,., Dq) 9Rv
which are regular at
q
is a regular differential of
valuation Notation:
of
v.
It is clear that the derivations which are regular at module over
v
K/k
v.
A differential
if it is regular at each
K/k. regular at
v).
Using this notation, we have ~ (~q) c R v. q v Let V be a variety such that k(V) = K. Def. 8.5: Let D
Def. 8.6:
P be a point of V is r_egular at
P
if
integral closure of
~
D
a derivation of
D~p c ~p
in K. P ~jq is re~lar at
A differential for all
and
DI, ..., Dq
~
q
P
if
~p
iB the
a~q(Dl, ..., Dq) a
which are regular at
tOq is regular on V Prop. 8.7: A differential
where
K/k. Then
P.
A differential
if it is regular at each point of V. is regular on V
if either one of the
following conditions is satisfied: (I)
~Jq is regular at each rational point of V.
(2)
~q
is regular at each irreducible (r-l)-dimensional
subvariety Proof. Assume (I) and let ization of
P
over
k;
~P = ( ~P' )S where
~p,.
Let
element
f ~ S
of V.
P be any point of V.
then
i.e.
D
~-
~p S
fD
P'
be a special-
is a ring of fractions of
~p,,
is a multiplicatively closed subset of
be a derivation which is regular at such that
Let
is regular at
P'.
P.
There exists an
Hence if
DI, ..., Dq
P
-2"H are regular at P, we can find
f~S
for
i = l, ..., q. Now every point
P';
and since k
such that fD i is regular at
P has an algebraic specialization
is algebraically closed,
is regular at P', we have
~Jq (s
P'
is rational. Since
..., fDq)
P~V
is regular at P.
and assume DI, ..., Dq
are regular at P.
Then D i ~ P ~_ ~p,
i = l, ..., q. We note that if ~ i s
k(V)
~, then D ~ s
and if D o~
closed subset of ~ Df
Therefore
s
P,
~md so
Let A = Wq(Dl, ..., Dq) where ~
S
where
S
DI, ..., Dq
~, hCs,
then
are regular
at all
(D1, ..., Dq) E ~ for all such F . q P and assume A ~ ~p. We know ~ p = ~ ( ~ p ) ~
x~;n8 through the m~nimal (rank l) prime ideals of
there exists a ~
~
such that A ~
9 Then
A# ~
(~p)A
. Let
~
~. Hence P be the center of the
which is a contradiction. Hence A~ F
and
F
~
i
w luation
any subring of
is a multiplicatively
; for if f = g/h ~ ~rs, g, h ~
(I/h~)(hDg - g D h ) ~ ~ .
which contain
c_ ~
q
= fq~Jq (DI,...,Dq) ~ p ,
Iience ~Jq(D1, ..., Dq)~ ~p , showing that ~ q Assume (2). Let
P'
P
~j is regular at P. q Now, let P be any irreducible (r-1)-dimensional subvariety of V
such that
Pg P.
It is immediately seen that the integral closure
of the local ring
(~r(V/k) is the ring of quotients of
to the ultiplicative system
are regul~r at ~-.
Hence
- #,
where 2
6jQ(DI, ..., Dq)~ ~ .
~p
with respec~t
is the pr e ideaZ
Since the intersection of
i
all the rings ~neorem 8.8"
~ is ~ , ~ F P q If
~q
is regular at
P.
is a regular differential of the field
K/k, then
OJq is regular on V I and conversely, provided V
is non-
singular. Remark- We shall assume q=l. The proof for larger values of q same (but with more indices ).
is the
-28-
Proof: Assume ~ d~m~
= r-l,
is a regular differential of
let ~
Pc
V
here
(V/k) and let ~ be the integral closure of F . There are a finite number of prime divisors Vl, ..., vh of K/k
with center ~ .
=
K/k. Let
~
There exist exactly
all of which are maximal, and
h ~
prime ideals i = Rvi'
~l' "'" ~ h h ~ = i=l ~ R v4
in
Let D
be a derivation which is regular at ~ , Then D ~f c ~ and so h DRv. ~ Rv.. Hence ~ ( D ) ~ Rvi ~. Therefore tJ is regular at [-~ i
i
and Prop. 8.7(2) shows that CO is regular on V. Now assume ar~ valuation of
V
is non-singular, and cO is regular on V.
K/k, and let W
an irreducible subvarlety of V).
be the center of
v
Fix a rational point
on V P
Let
v be
(hence W
on W
ar~
uniformizing parameters tl, ..., tr of P on V. We can write r = ~ Aidti where the A i are in K. Indeed, the regularity of i=l implies the A i are in ~ p in view of Theorem 8.2. Let D be any r derivation which is regular at v. We can write D = i--~ Bi ~t--~ " Since
ti ~ P
)
~ ~CRv,
we have
(D) = i= ~
Dt i -- B i~ R v.
BiA i E R v.
Therefore
Hence, since ~
is regular at
v.
is
-299.
The canonical system on a variety V.
Def. 9.1:
If W
is a simple irreducible subvariety of V/k,
~i' "''' ~r ~ =
~ W (V/k) (r = dim V), then the
uniformizing coordinates of W (a) [ ~I' ""'
and ~i
are
if
~r ~ is a separating transcemdsnce basis of K/k
(b) a_Z_ Let W
be any simple irreducible subvariety of V/k.
Then it is
easily seen that there exist sets of uniformizing coordinates. P(W
where
P
is a rational simple point of V.
be a set of unlformizing parameters cf (a) holds; and since
~
P.
For, let
Let { tl' "''' % }
Then by Theorem 8.2 we know
are regular at
P,
they are regular at W.
ti
An alternative method is as follows: sentative of V
such that W
i.e., W ~ V - V a.
Let
let V a be an affine repre-
has an s/fine representative W a
k[V a] = k[Xl, ..., Xn] , xis
[fl(X), ..., fN(X)) be a basis of
in Vap
and let
l(Va). Consider the Jacobian
...,
9W
is simple
on V
if and only if this matrix has
a(x l, ...,xn) rank n-r
on W,
hence we can assume
~(fl' "''' fn-r ) # 0
on W.
~(Xr+l, ...,Xn) Thus ~ Xl, ..., X r ~ is easily seen to be a set of uniformizing coordinates for W. Prop. 9.2:
If
~I' "''' Jr
if
71, ...,~r ,{~,
coordinates of W
are uniformizing coordinates of W then
7 1 , ""' ~ r
if and only if
and
are uniformizing
a(ql~...~r) ~ ~"W" a( )
Proof:
Assume the
~i
~(~)
3(El
and (b)implies
= I,
are uniformizing coordinates. ~(~)
~(~)
Then
~.
Hence
ac )
is a unit in ~ . Let
Di = zr A
the Aij
~
,
i = I, ..., r, AijE K.
so that ~ e syst~n of linear equations
Regarding these as equations for the which is a unit in ~
A
Let to
V,
~
be a prime divisor of
and let W
~ ~" k[z l , Let ~
Wa
...,
z m]
contained in
field).
k(~) = k(g)
We call ~
on V
(hence
R
9
d i m W = r-l).
such that W
V a . Let
(because
S ....; ~ c ~
of the first kind with respect
of V
(Zl, ..., Zm)
a (~)
AijE~.
c 8 ~ since
R = k~Va],
be t h e i n t e g r a l c l o s u r e of
be the locus of
variety, and
Di ~ K/k
Va
holds.
the determinant is
be the center of ~
Choose an affine representative re~esentative
D i ~ j = 5ij
o Hence we can find such
This proves (a) of Def. 9.1. Clearly
We want to choose
over
and let
R in its quotient field~ k.
and ~
a normalization of V a.
has an affine
Then ~
is an affine
have the same quotient We note that ~
is
detelu~ined only up to a biregular transformation since we only know ~ ~' coordinate ring. Let W W
be the center of ~
on V.
Clearly
is represented by a minimal prime ideal in R.
ring of ~
is
because
Hence the valuation
C~W(~/k ). By a set of uniformizin~ coordinates of
we m~an any set of uniformizing coordinates of W Let
dim W = r-1
~Or be an r-fold differential of
on V.
k(V)/k, ~Or # O; let
be a prime divisor of the first kind with respect to be uniformizing coordinates of
~
.
Then
d~l
V;
and let
"'" d ~r ~ O,
~I"'"
and we
~r
-316Jr = A d ~ l .. . d~r
can write Let bJr
v~
__at ~
where A6 K.
be the valuation defined by (notation" v~ (UOr))
to be
~
. We define the order of
v~(A).
To see that
v~ ( GOr)
is independent of the choice of uniformizing coordinates, le~, 171' ""' ~r be another set. Then
LJ
= B d ~ l "'" d ~ r
and
B = A "I~ ( ~ ) i
9
~( ~)I ) Hence v~ (A) = ~ (B) ~ince v~ (I ~~((~. ) ) = o (by ~op. 9.2, this r
Jacobian determinant is a unit in the quotient ring Prop. 9.3"
Let
Or
~Y(~/k)). Then
be an r-fold differential of k(V)/k~ cJr ~ O.
v~, (~Jr) = o for all but a finite number of prime divisors of the first kind with respect to V. Proof: Since
Let V a = V - (V(TH) V ~ H
H
V
by its affine representative
R = k[Va~ = k[Xl, ... Xn~
be the integral closure of i = I, ..., r.
R,
~ z~/~x i
s
Xl, ..., x r
= A~
and consider
Let ~ = k[Zl, ..., z ~
~ zp/~xi,
~ = I, ..., m;
, whence
for all
i
The divisor
then for all ~ v~ (C~r) = ~
and all ~
~ d xi R ~ C R~
are uniformizing coordinates of
... dXr, (A~ K),
~ zp/@x i
are finite at all the remaining prime
~ z~/~xi~ R~
~ax----~CR~
we have excluded, we have Def. 9.4:
k(V)/k.
and so we can discard them. Hence
. Therefore
This implies that and so
Va .
There is only a finite number of prime divisors which are
( ~ = I, ..., m; i = I, ..., r) divisors ~
H.
where we can assume that ~ Xl, ..., Xr~
is a separating transcendence basis of
poles of the
is a hyperplane such that V ~
contains only a finite number of prime cycles, we may replace
in the proof the variety Let
where
~
for
.
i = I, ..., r;
. Hence, if
outside the two finite sets
(A) which proves the proposition.
(~r) = ~ v~ (OJr) 9 ~ ,
6jr ~ O,
where ~
runs
the set of all the prime divisors of the first kind with respect to V,
is called the divisor of the differential
6~ 9 r
-32A canonical divisor on V ~r'
O3r # O.
Let
~ , W'
is the divisor of an r-fold differential
be r-fold differentials,
%
# O, ~' # O.
r
Then CJ' = B ~ r
(~Jr)
where
r
BE k(V). IIence (co') = (B) + (Or), i.e.
r
and
r
(GJ')
are linearly equivalent.
Thus all canonical divisors
r
belong to one and the same divisor class, and, furthermore, it is clear that any divisor linearly equivalent to a canonical divisor is canonical. This divisor class is called the canonical divisor class. canonical divisor, then
If
K
is ar~
lE } is the canonical system on V.
An immediate consequence of our definitions is Prop. 9.5: Let tion,
A differential
~r
is regular on
V
if and only if
(CJr) >- O.
c~ be a regular differential, and let K = (~r). By definir is an effective divisor. Conversely, if we start with an
K
effective canonical divisor
K,
then
~J
is determined to within a
r
non-zero constant factor in
k.
Let
L
of the regular differentials on V, L -~0}*
IK# , where
car
as a projective space. Since fact,
be the vector space (over k)
of degree
car),
%
dim (K~
/ o.
< |
dim L -- I + dim ~KJ . We denote
r.
Thus we can consider we have
dim L by
pg(V)
the geometric .genus of the variety V.
p g(V)
linearly independent differentials of degree
on V.
Clearly,
If
V
pg(V)
We have a mapping
dim L < =:
IKI
in
p g(V), and call
Hence
V r
carries exactly which are regular
is al~ays => O.
is biregularly equivalent to
V',
the
pg(V) = pg(V').
This is not necessarily true for birationally equivalent varieties. However, if V
is non-singular and
because any differential of degree on
k(V)
and hence on V'.
k(V) = k(V'), r
the
p g(V) -< pg(V'),
which is regular on V
This proves
is regular
-33prooF. 9.6: Any two non-singular models V,V' field ~ (= k(V) = k(V'))
of a given function
have the same geometric genus.
Let ~/k be a function field where t.d. Z/k = r. We define the gepmetric genus of ~/k, p (Z/k), to be the minimumvalue of
p g(V)
g
~here V
is any proJective model of ~/k.
Theorem 9.7:
If V/k
is a normal variety, then the differentials of a
given degree
q which are regular on V
form a finite-
dimensional vector space over k. Proof: We ~all assume
q = I
since the proof is the same for any q.
Fix a separating transcendence basis { ~i' "'"
~r ~ of k(V)/k,
and let &2 be any differential of degree I which is regular on V. we can write
~ = Ald ~I + "'" + Ard ~r"
Then
Let ~- be a prime divisorial
cycle. Case i: Assume the
~i
are uniformizing coordinates at r
.
Since C) is regular (in particular, regular at ~ ) the A i must be regular at ~ . Case 2: .
Let
,
~I' "'"
Hence
vp(Ai)~O
Assume the ~r
~ i are not uniformizing coordinates at ~
be unifcrmizing coordinates at ~ .
d~i"" Since the
~i
for i = I, ..., r.
d~r =
.
Then
i ~(~I~ "'" ~r)id d~r. '~(~i' "'" ~r ) NI""
are uniformizing coordinates, we have
v (d~l... d ~ r ) = v
( B ( ~ ) J = coefficient of ~
in the divisor
r d~r). Since the
~ i are not uniformizing coordinates of
is infinite at ~
or
~
is a component of
cycles r .
Let
eihher some
~i
(d~ I ... d ~ r )" Thus there
are only a finite number of prime divisoria! cycles are not uniformizing coordinates of ~ .
~,
~Ich that the
dT be the set of these
~i
Let
~
~- 9 Then
6q~, and fix uniformizing coordinates ~=
r B Ai = ~ j=lJ
B l d ~ l + "'" + BrdBr
~J ~i
i = I, ..., r.
orders of the
where
B1, "" .' Br 6 O/r " Then
This gives us a lower bound fur the
..., v
(...,
r
a~
Denote the right-h~d side of (*) by s(t--). si o sd'-i).
Hence each
Ai
Z o "•
, 0
CJ
.
Let C~ ~ I--1, . . . ,
~tl--i"
(*)
l--h~ ,
We have sho~n ~ t
(A•247 k,
-~0,
and
.
Trace of a differential.
Let W and
De~e
h
, ...), ...
varies in a finite-dimensional vector space over
h e n c e so does
w
at
A i, namely
vp(A i)-
a ~ let
~ I' "" 9r
be a subvariety of V;
~ = k(V). Let ~
at W;
then ~ Y
and let ~
= ~w(V/k), ~
= ~w(V/k),
be the set of all derivations which are regular
is an ~-module.
Let ~
=~DE~T~D
~
c ~.
W
Def. 10.1s
If
D g~ W
we define the trace
the derivation
~
where ~?@ ~ , n ~ If defined.
~ ~ ~,
of
k(W)
and
f
D
n
W
such that ~ ( ~ ) =
= TrW I ? ( ~
then D ~ g a ~
T~
Trw(D ~)
k(W) = ~//~).
TrW ~
It is immediately seen that D
to be
= 0.
Hence
E
is well-
is indeed a derivation of
k(W)/k. Prop. I0.2"
If W of
Proof: Let trivial. ~I' "'"
is a simple subvariety of V,
k(W)/k
s = dimW.
is the trace of some derivation in ~ . If
Hence we may assume ~r
then every derivation
s = O, s > O.
then W
is a point and the result is
Fix uniformizing coordinates
and uniformi~ing parameters
tl, ..., tr_s
of W
on
-35ti
r-s
V/k
The quantities
(hence /M~ = Z ~ft i). i=l
n2 = I, ..., r,
,
~t i
~) = I, ..., r;
and let
i = I, ..., r-s~
-
Let T~W W ~ j
belong to ~ .
i = I, ..., r-s;
~
Z f ~ ~ k(W),
=
~i = (Zil' "''' ~" )'
i = I,
mr
eee~
m
r-s.
Let
r-s-~
be the number of the vectors
independent over e2
k(~)
(hence ~ > 0).
(0,I,0,..., 0), ..., e
Let
Z.
which are linearly
I
e I = (I,0, ..., 0),
= (0, ..., 0,I)
We can assume that
r
el, ..., e
span a space complementary to that spanned by the r
We can find
L
=
~
a
a
a)=l
a--I,
X~,
..., s + ~r, aa
ek(W),
a~
such that
Let
aa~ 6 ~
have
La(e~) = 8a~,
~-I,
..., s + e-
~(zt)
i
,,.j
= o
W-trace
,
~a 9'
= I~
r
-
s
and consider the derivations
r
D~ ~ z a~ a ~) =l
all
a.
~ ~
, ~
We have
~ o 1, =
Dat i
...,
~ a
~ti
12~I a ~
-aa~i~ = La(~i) = O. ~=I i = I, ..., r - s~ and so
s § ~
~,~
Therefore D~ ~
.
~ ~
Cl~r~
"
for
for
~
Hence
D~t i ~
~ ~.
D" E ~ W a
Trw(Dat i) =
a = i, ...~ s + O-
This shows
DI,- ..., D ~
a
~
g
Now consider the traces
D~
= 6a
D
(rood ~ ~.
Cpnsequently
then .....
),
Hence O" = O.
~ ~till
Joe.,
of the ~
2
DI, ..., D + ~ _
D~
= 8a
on
W.
~ = I, ..., s + CF-,
where
are linearly independent over
k(W).
r-s
has (maximum possible ) rank
It follows that we can replace
W
We have
Thus we have shown that if /M4 =
We may therefore assume that
6--~~ .
s+C[
r-s
s+i = ti,
~ ~ti, i=l
r-s
on
W.
of the
~
ts
by the
i = i, ..., r - s.
Let
t~s.
-36De = ~
, m = I, ..., s.
= I, ..., s;
D~ ~ c ~ , and let
and since the
~ = I, ..., s. ~
D ti = 0
ti
for
E~
i = i, ..., r - s~
form a basis of A~%,
Therefore
DI, ~
= TrW ~ 9 Then, by definition,
Therefore the k(W)/k,
Then
.:~W
D ~a~
form a basis of the space ~
we have Let
= 6 ~
(W)
for
D~
%~
= l,..o,s.
of all derivations of
and this proves the proposition. If W
Prop. 10.3:
is a simple subvariety of
(a) ~ W
is a free r-dimensional
(b)
W
(c) ~ W / a ~ ~ W Proof:
Let
D 6~W
if, sn%d only if,
I' " " "
~r
V/k
of dimension
~-module ( ~ =
s,
then
~fw(V/k)~
and
is a free s-dimensional
~/~
-module.
be uniformizing coordinates of W. Then r O = 7 Ai , with A 6 ~ since I
D
f
A..l Therefore ~i
~ ~ i,...,
c~
I form an CT -basis of
are linearly independent
and the module is free because the over
k(V).
~,
This proves (a).
(b) is obvious. Let where the
~I ti
be chosen so that
~s+i
are uniformizing parameters of W. a
S
writs
~r
'''"
D =
EAi-~ i=l if~ and only if, (1)
~,Bj ~ ~
(2)
Dtj ~ .
i +
r-s Y Bj
j=l
Then
t i , i = i, ..., r-s,
Then, if
D ~c
~
~
D ~W, and
D~w
we can c /v~
~j
, i = l, ..., s; j = l, ..., r - s~ and
(1) and (2) are equivalent to
Bj 6 ~, J = l, ..., r - s.
This proves (c).
We now give a characterization of uniformizing coordinatest Prop. I O . ~
Let W
be a simple s-dimensional subvariety of
~I' "'"
~r ~=
coordinates of W
~W
(V/k).
on
V
Then the
~i
V/k,
and let
are uniformizing
if, and only if, the following two
-3?conditions are satisfied. (a)
k(W)
is a separable algebraic extension of
where (b)
~ l 9
The ring
]., . . . ,
i, i
--
k[ ~I, "'" ~ r ]
parameters of W
k(~l, ..., ~r )
r.
contains a set of uniformizing
V/k.
on
Furthermore, (b) is equivalent to (b') Let ~ q of
A~q
= NWw(V/k).
Then a~q f] k[ ~ ] contains a basis
.
Proof: Assume (a) and (b). Fix uniformizing coordinates ~ I' "'" ~ r of W
on V/k.
By virtue of Prop. 9.2 it suffices to show that
d('~)
8(~)
where not all the
A.l are in ~
,
say A l ~ ,
r >2. A
such that
~i~
r
Let
D =
F
Aj
Since the ~ j are uniformizing coordinates, we have D
is regular on W.
D(k[~])~4h~. and so
D
has a trace
on
~ = Try.
D~
k(~l, ..., ~r ) and hence hence ~i
~ ~I
i.e.~
it fellows that
c /~4. Therefore
D~iE,Y~ implies
This shows that the
Dg~W
~ ~i = O.
is trivial on = ~#
0
,
This shows k(W)
by
and this is a
are uniformizing coordinates of
V.
Now assume
~ I' "'" ~ r
are uniformizing coordinates of W.
~how (a) we need only prove that if that ~ = 0 Then
i = I, ...~ r,
On the other hand, D ~ I = ~ '
contradiction. W
D~i 4~,
Hence, by (b), we have
is trivial on (a).
Since
D ~ ~W'
D~i6~.
we have
D ~r
on
k(~), Let
D
then ~ = O
Ai = D ~ i ~
on
then
/v~ which shows that
is a derivation of
k(W)/k
To such
k(~). Let ~ = Try, D g W r D D = E A. Since Ai~ ~
~ = 0
on
k(W).
-38-
Since (a) holds, we can assume ~ ~ I ' transcendence basis of
k(W)/k.
minimal polynomial of
s+i
Let
over
"'''
fi ( fl, ..., ~s+i-l' X) k(
I' ""'
By clearing denominators we may assume that s~l
arguments (of course, Let
since
ti = f(~l' "'"
fi(~l
, ...,
We shall show that precisely, if A i~ ~ ti
9
fi
for
~s§
~s+i )"
~s+i ) : O, we have tl, ..., tr_s
is a polynomaal in all its
am~,
r-s.
are uniformizing parameters of W.
O'I/i~). We know
More
we shall show that
(this means that the ~
2-residues of the ~
Eam.
Hence
) ~- ~ .
j:l separable algebraic over
~-s~"
ti~k[~]~
~ ] / ~ N~, i = I, , . . ,
~ = Alt I +...+ A r . e % . s ~ ~ AiE ~ , i = I, ..., r-s
for i : I, ..., r-s.
Then clearly
tick[
r-s #A. -Z Stj=A
Therefore
fi
~s+i-l)
be the
now need not be monic).
are linearly independent over
'
~s ~ is a separating
Since
k(W)
is
s
k(~l, " ' "
~S ) we have
.
~r
In a similar fashion we see that A i E ~
for
i = I, ..., r-s.
Special case. Let
s = r - I, i.e., let W
cycle. Then t.d. k(W)/k = r - I. uniformizing coordinates of W. r - I
Let
{~i, "'"
Then among the
algebraically independent elements.
are multiples of one irreducible polynomial denote by
t.
Since
k[~] F T ~
~r~
be a set of
~i' i ~ i, ..., r,
Thus all elements of f(~ I' "'"
are
k[~ ] / 7 ~
~ r ) which we
contains a basis of /vm ,
must be a set of uniformizing parameters, i.e., ~ Hence
be a prime divisorial
the set
is a principal ideal.
Vw(t) = I. Let W
I~I' "'"
~t
be a simple s-dimensional subvariety of
V,
~r ~ be a set of uniformizing coordinates of W
and let such that
-39{~
is a set of uniformizing parameters of W.
@
S+I'
can write an
=
~ ~
~
~/~-module, i.e., a
we see that ~
~
+ a~
Then we
. We have seen that ~ W
k(W)-mcdule.
By Prop. 10.2 and Prop. 10.3(c),
can be identified with the space ~ (W)
of
W
derivations of Let GO
q
k(W)/k. be a q-fold differential, and assume &)q
is regular at W.
~hen we have the followingt If DI, ..., Dq
Remark:
is in / ~ W '
are regular at W
then C~q(D1, ..., Dq)~ ~ .
To see this we need only expand aS
O~q(D1, ..., Dq).
Dl, ..., DqgO~W, cosets of DI, Def. IO.5:
If
G3q in terms of
and lock
d ~l' "'" d ~ r
I~ imm3diate consequence of the remark is ~ if I
then TrwaJq(D1, ..., Dq)
...,
Let
and at least one of the D i
Dq,
depends only on the
i.e., on the traces ~i = Tr~i' i = i,
D Q@~
q.
(~)q be a q-fold differential which is rega]ar at W. q < s, we define m
T r w % ( E I, ..., ~q) = Trw~q(D1, ..., Dq) q>s, we fix define
q-s derivations D~ ' "'" D ,.. 9 ,D t Tr q-s gO by W
Dt q-s
in ~
and we
W
q
D' "' q-s Gjq)(~,...,~s) where
DiE
, T
i =
=
Trw((A~q(DI ~
,..j
DS~
D I ~o.., D'q-S ))
i = I, ..., s.
We now list some simple properties of the trace of a differential:
(2)
(3)
--
If ~ w ( V / k ) ,
then
Trwd ~ = d~
where ~
= TrW ~ .
-40Let
I~l, ..., ~r ) be a set of uniformizing coordinates of W,
and let ~)q
l'<Jl<J2<"'<Jq
)
Jl
Jq A(j)
(4) If q <_ s, then Trw~% = 7 ~(j ) d~jl ... d ~jq where A(j) = TrwA(j ) and
~Ji = TrW ~Ji"
We turn now to the case q > s.
It is clear that if at least one of
the derivations Dl,' ..., D'q~.s ------is in ~ ~ For, if D~6~9 W' ~
' then ~ D ~ ' ' " ~ D q ' s c j q
= O.
then, since ~ ~
is a free s-dimensional W ~/~-modulo, we have D1, ..., De, D 1' linearly dependent (mod A~q~W) over ~ / ~ . Therefore %(D1, ..., De, D~, ..., D'q_s)6 ~ and so Di,. "" D'q-sr Tr =0. W q Let [tl, ..., %-s ~ be a set of uniformi~ing parameters of W, and complete it to a set of W where
{~I' "''' ~r I of uniformizing coordinates r ~s+i = ti' i = I, ..., r-s. Then ~ W = 7 ~
o 8 7S ~ ..... and o~/W = ~=I ~a +~
~W
. Hence any trace of W will be linearly
dependent (over k(W)) on
ql+S'...,
iq_s+
w
where
(*)
I < i I < i2 < ... < iq_s ~ r - s. We write (*) as
til' .. "'tiq_ s ; ~ I ' ' ' " If
S
~Oq = 7 A(j)d ~jl...d~jq,
tij; ~k _ TrW C~q = A1,"
if+
I < Jl < J2 < "'" < Jq -< r , then
iq
+ S
Special case: Since
q-s = r-s,
q = r. we have
If
q = r, then
6 0 = Ad~l... d ~ r . d
TrW
where we note that Ad ~ i " "
d% s
I""
s
is not indeoendent~ of the choice of
~I' "'" ~s" Special Case: this case we have
q = r, s = r - I~ i.e, TrW
t;~l,..., ~
independent of the choice of the differential of degree
r
fotmizing parameter of
W,
Let curve on
F F.
[-
is a prime cycle.
- _
r - l ~ o = Ad~l... ~i"
Thus if
which is regular on then
Tr~ ~
Let 00
[- .
d~r_l
W
and if
t
(6J). Finally, let ((4) + r" _ (t)
~
be an irreducible
t
c
is a cycle which does not have
tO1
s(F)
is. independent of OJ and
t.
be another differential satisfying the same conditions as
We can write
COl= A~
where
Ae~,
A4~.
Choose
is a set of uniformizing coordinates of Then
k(r').
[ ( ~ ) + [- - (t)] - (Tr t UJ ). C
~e shall show first that Let
F-
be a uniformizing para-
Tr [(&)) + [- - (t)] is a divisor in rwe can consider the divisor (Tr t GJ). Let r-
s(~-) = Tr
Tr boa = B d ~ r
is a
is well-defined.
be a differential of dogree t~o, and assume
Then
Tr t aj ~ O, r
aJ
is a uni-
as a component, and
Since
Ln
and this is
dim W = ~-l, if
be a non-singular surface, and let
is not a component of meter of
W
and T r ~ l
Tr [(~l) + [- - (t)] - (Tr t
on
= ~ d~ .
F,
$
so that
and let
~
UJ 9 {$,tl
- Bd~dt.
Hence
(t)] -(i)= (Tr t ~ ) ~ l ) = (X) + ~
r [(~) + F
-
is(C-).
Let
~
We have
be another uniformizing parameter of tI = Bt
where
B
is a unit in
9 .
W
and let Then
~
be as before.
. . . .
I ~ (t, ~ )'
B
-- t
at
which is a unit in ~" . Hence coordinates of W. Hence
Let
Tr tl u 3 - ~ l d ~
is also a set of uniformizing
CJ = Ad~dt = Ald~dtl;
Tr t uo = ~ d 7 = (B + ~ t ) Ald~ = B ~ d ~ F at s(F-) is independent of t as well as of •
it is clear now that
The divisor
s([-)
The divisor
s([-) = 0
singular points. of
k(~-)
F
Let
if and only if
%hen
s(~)p
in its integral closure. Proof: Assume
e~ationof ~p(r'/k).
P
F
is a simple point of
P
is simple,
Since
~
where
~
s([-)
9
and let
x = 0 be a local
Consider ~ p ( F / k ) / ~ p ( F ~ = is a regular local ring.
is a principal ideal.
x = 0
since
Since
Tr x dxdy = ~
at
s(~-)
Then
- (t))p = O,
F
at
is a basis
P~
t = x
and so
and
at
P
P where
x = 0
is a uniformlzing parameter
is independent of 6J ,
(63)p = 0
and
Then
x,y
Pc
P.
be a set of uniformizlng parameters at
is regular at F . ((~) + F
y~ ~s
it follows that ~x,y~
is a local equation of ~-
is a local equation of ~" on F.
Let
Therefore { x,y I is a set of nnifoz~izing parameters at
Let {x~y]
of ~-
~p(~-/k)
(We shall not p~-ove (c)).
This shows that we can always choose uniformizing parameters such that
7
if and only if 7
is the [--residue of some element
is a basis of ~p(~'/k),
of f~p(F/k).
~-,
~p(F/k)
Hence /V~p(~-/k) = 4~p(F/k)/~p(F/k~ ~p(~-/k) = ~ )
has no
is the conductor of
at P (hence X6~p(F/k)).
Since
~-
More precisely 3 a prime divisor
is a component of
Ps
.
has even degree.
is centered at a singular point of F (c)
.
is effective, and its local
component divisor at each point of (b)
A = (B + @---~Bt~A ~t j 1.
then
and
F
Theorem 10.6: (a)
t~ ,tl~
(t)p = P .
let 6J - cLx&y, which Therefore
Tr ( ( ~ ) + r- - (t))p = o. Furthermore, t" since y is a u n i f o ~ z i n g parameter of P ~,n
F,
we have
(dy)p = O.
(Trx@.~ )p = O. ls ( ~ ) ~ 0.
has no singularities, then Assume
P
is a singular point of
u_niformlzing parameters at of
~"
at
P.
This shows that if
Thus
P
on
F,
~.
Let { x,y}
and let
t = 0
Again we can let 03 = dxdy,
be a set
be a local equation
and so we have as before
(t)p = [" and ( ~ ) p = O. Hence - (Tr tr
Tr ((~s) + [" . (t))p- (TrFt ~O)p = fbe the set of prime divisors Y of k ( F ) which
dxdy)p. Let B
are centered at
P.
Let ~
~ ~r t dxdy I r
(~)p ,. z
then
vy(~U) ~ Y
@
Ts
To prove (a) we must shcw:
(1) vy(~) < 0 for all Y6B, (2)
~
vy(a ) is even,
Y6B We adopt the following notations: and
and
(~=
dp(F/k), M - ~p(F/k),
~' ~ ~ [-(F/k
/~A = ~p~/k).__ We can consider
exist uniqae elements lcok on M/M 2
M/M 2 c,d
as a vector space over such that
~
k.
For ~ ~ M,
- (cx + dy)~I~ 2.
as the two~limensicnal set of linear forms
there
Thus we can cX + dye
The
elements of the corresponding one-dimensional projective space are definable by equations of the form
cX + dY = 0.
projective l~ne directions at Let ~ ( M ; and
~ ~ M u+l.
P.
there exists an inbeger Hence we can write
where
ai ~
~ , i = O, .o., n,
dezote
the
M~resid~e of
+ ~nYn
We call the elements cf this
~
mud not all the -
a i ~ k).
ai
of ~
%hen
biGM
f~
are in
We call
-
aoX
n
M~ + ~i X
Let ~i n-
~
+ ...
. A well-known characteristic property of
uniformizing parameters cf a regular local ring is: + bn yn ~ 0,
such that ~ ( M n
~ = acxn 9 a W l'l y + ... + an yn
a i (hence
the leading fo~m of
n, n _> !,
aY~
i.
if
boxn + blxn'ly + ...
Eence we see that %he leading form
is independent of the representation
~ = aox~ + s~x n'l y + ... + a n ~
If
oX + dY is a factor of the leading form of ~
, then the direction
oX + dY is said to be_tangent to the cycle (~)p Let ~ Since
T
and ~
by the F--traces of x
is centered at
P, we have
y
x
and y - cx where
and let
v T ~ ) > 0 and v ~ ) Q
Furthermore, since we can always repl~ ce x parameters
and y,
> O.
and y by the uniformizing
c is a unit in
~
(i.eo~ change x
by a non-singular linear transformation), we can assume
96 M s+l where We C~-~write Then Y
s.
Therefore
is the leading form of
t = aoxS + ~ x s ' l + .., + asyS where
will be a factor of the leading form of t ~t,
is a unit in
~p(F/k) = ~ I d t ,
t6~,
t " (CoXS + OlXS-~ + ... + CS~)
Coxs + clXS'~ + ... + csyS
residues mod
we have ~c~ s + ~ l ~ s ' ~ +
ai 6 ~ , if
t.
i = 0,..., n.
ao~ M.
.,o +~s~ s= O.
Takir~ If ~o
then v7(~o) = O. But this yields an
obvious contradiction with the fact that
v7(~) < v7(~) . Hence
aos
Y = 0 is tangent to ~-.
not tangent to if
m,~E ~
Hence
~I
fact, ~ I Let
(~)].
where
A well-known result for regular local rings is:
m~
, s~+l
~Mg ' ~Mg+l
is a ring. Clearly ~ c ~
then
~I.
(a)
~E Mn+l 3 or
(b)
~ M n+l, and Y = 0
i. We shall now show that, in
Then
~
is a non-unit if and only if
is tangent to
(~).
It follows that the non-units form an ideal M I in It is clear that x ~ . 1
g+h+l.
is a regular local r~ng of dimension two. ~ = ~/~
and
vT(~) < vu
Then we assert that Y = 0 is tangent to ~- . For we have t ~ M s+l for some integer
7EB.
~I"
Let Yl = y/x~ then Yl
but even ~n M I. We claim that x
and Yl
is not only in
form a basis of ~ .
and
-45Let
~ - ~/~6~.
lqe can write
where bee M. Since M = ~(x,y), ~xn~l(X'Yl)
~ = b~
we see that ~ M = Olx. Hence 1 ~ = xn~l where ~16 ~l(X'Yl >~
' and so we have
We can write ~q = ( ~ ) x n. Since ~ M to (YI)' we have ~ / x n 6 ~I"
+blXn'~ + ... + b n ~
a
and Y = 0 is not tangent
Therefore ~
= ~ i xn is a unit in
I"
Hence E/~ 6 q ( x , Y I) which shows that x and Yl generate N I. Consider
ff[yl ] ~ ~1'
and let H = lao + alY + . . ~ a~ml ais
ao~ M) . It is clear that H is a multipllcatlvely closed s;~t and that ~ i = ~[Yl]H 9 Since ~[yl ] is noetherian, it follows that ~ I noetherian and hence ~ l
is a local ring. To prove ~ l
is
is regular
we must show dimk MI/M12 = 2. It is clear that d ~ To show that ~m~ ~ 2
Ml~12_<2
= 2,
where ~ , ~l,qEMU;
then A,B 6 ~ .
We can write i = ~ / ~
q $ M n+l and Y = 0 is not tangent to
(q).
Then
~x 2 + ~I y = O. Since % x 26M n+2, y E N
that
~i E M n+l, Therefore B is a non-unit in
Furthermore, since Y = 0 is a non-unit in
~l"
generate M1.
it suffices to prove the following assertien:
if I~ + ByI = 0 where A, B g ~ I , B = ~l/q
since x and Yl
is tangent to
This shows d ~
(~),
MI~2
and y ~ M 2, we see
, i,eo, B~M]. we see that ~ / ~
=A
= 2, and therefore
I
is a regular local ring. Let
~ = ~t;
then D 6 M n, ~ M
then ~" = ~
and hence ~ C
~ . Let
n+l and Y = 0 is not tangent to
~/~e~l$
(~]). This means
~ 0 (rood t) since Y = 0 is tangent to t. Hence ~/~ 6 ~sm.d 1 Let ~l
~ l = / ~ q ~ ~l where we recall that ~ 9 = ~(F/k).
is a prime ideal in ~l'
and /~4/q~ = ~ = ~ t .
This yields the following sequence of inclusions ~ = ~ Therefore
~ = l~l"
Then
Hence ~ l / ~ " C~
~ (.~=
-~6-
We have defined t~M s
and
where
tl~ C~I
Since
XlSh~ ~I
~ = ~t.
t ~ M s+l.
Viewing
~ i ~ ~Itl . Clear~y
Since ~ g ~ , ~d
t
and ~xl)Yl} and
There is an integer as in ~ ,
#Itl =
such that
we can write
t = xlst I
is a set of uniformizing parameters of
xI~])
we have
~[tl ~ ~l"
tl~l.
~t
we see that ~ ~ C t.
~i = v/~n
~.
We shall show that
~ = g / ~ s bl ~ ~
Th~ef=e ~'
~ = ~'t where ~ ' e ~ .
~l ~ ~I'
s
~e~.
is a u~t in
~,
~ ~#-s ~l where
and ~= :~l_ni~ltl. Hence ~ = ( ~ l / ~ l ) t l ~ ~itl , and so ~I" This shows that ~ = 0 is a local equation of ~- in ~I"
What we have described above is the effect, in a given direction at
~,
of a so-called "locally quadratic transformation"
center
P.
A full (global) description of
Let
Yo' " ' " Yn
general point is the point
A
of
F/k,
At
k.
Then
pondence with general point T t F-, F I
of
F
with
is as follows"
onto
F'
A'
whose homogeneous coordinates
(i)J) / (0,0),
and let
F'
be the
is a surface, and the algebraic corres-
(A,A')
F'.
P
Consider, in projective space of
the point
YiYj, 0 _~ i _~ J < n ,
over
F
and let us assume that the given point
dimension (1)(n+l)(n+2) - 2,
locus of
of
be (strictly) homogeneous coordinates of the
Yo = I, YI = "'" = Yn = O.
are the products
T
T
is a birational transformmtion
The following properties of
T
are easily
established: (I)
The total transform
rational curve (2)
T
F' - A'
(3)
A'
on
is biregular on
T[P~
of
P
is an irreducible, non-singular
F !. F - P
and
T -I
is regular on
FI
(biregular on
).
There is a cne-to-one correspondence between the directions of
at
P
and the points of
If
L = aX + bY = 0
F
A I.
is a direction at
P
and if
~'!p
is the local
-47ring
~p,(F'/k)
~;, = [ ~ / ~
of the corresponding point
I ~ , ~ E M u, ~ ~ M n+l
and
L
P'
of F ~(P's A'), then
is not a tangent to
(~)p~.
From the preceding "local analysis" it follows that F t is a non-sinEalar surface (in particular,
~ pt
is a regular local ring for each point
of A'), and that to the curve ~irreducible curve
there will correspond on F'
F ! whose intersections with
A'
B
is the set of valuations of
Let B = BltJB2LJ ...~B h where the B i
an
are the points
Pl' "'" Ph which cat~espond to the tangent directions of We recall that
p1
[-
at
P.
E(~-) w~th center
contain only valuations with the
s~me tangential direction, ~'e recall that { x~yS Let
P1 be one of the
h
is a set of uniformizing parameters of M.
points
Pi'
ar~ let (Xl, Yl}
be a set of
unifermizing parameters of M 1 where we may assume that x I = x y = XlY1.
Then dx dy = Xld~dy I.
v7(~) > 0
%
r
where ~ = Tr x. r Let
Therefore
tI
Trt r
YE B I. and
r
I~ence Tr t ~ r
= (i/~_is-l)Tr;l dXl~l.
Since we can assume v7(~l ) > 0.
v7(~) > vT(~) ,
there corresponds a
t1
r
)"
local, ring.
with uniformizing parameters xi,Yi.
~i
For each
obta~._u an expression analogous to (*) and ~ (for we may assume that
t F
it follows that
Therefore
vy(Tr; dxdy) = -(s-l)vy~) + vy(Tr TO each B i
and
v~-)> v~),
for all
§
(*)
(of the point
i = 152 , ..., h we is the same in each case v ~ B). Therefore
(tl i=l
Pi)
9 F '
P'
P.
We can assume that ~t,x~ of
F
on
F.
Then
Hence
~x
Tr
is a set of uniformizing coordinates
~ A~a ~ U 3x
and therefore either D t ~ ~t Tr ---- are not both zero. Let r" ~ Y
,t) and
~ x ~' Then
vT(Trr
c(Y;P) = 0 a non-negatlve integer since
~t
-~
ay
o~
1
~t ~ .
Hence
9y
c(7~P) = o if and only if s = 1 and P is therefore a simple point. We shall finish the proof of the theorem by induction on Assume
s > I.
We know
c(T; P) > c(Y~Pil) for %
s > 1.
c(Y;~).
c(7~P) _> (s-l)vT(~) + c(Y~Pi).
Therefore
Thus by the induction hypothesis
v ~ ( ~ r ~ ~Idyl) < o and so v~(~ t, ~dy) < O. The induction hypothesis also yields the fact that has even degree. dimk ~ / ~ ( t , x ) ... + a
By an earlier result we know that where
A
is
x = O.
where we must have
Is a basis of
~/~(t,x)
as
over
k.
deg (~)p ~ i(~-,A;P) ~
Now we can write
M.
t~ Z(Tr~dxidY• P
t = aoxS+alxS-~ §
Therefore the set
This shows that
1,y,~,
dimk~/OT(t,x) = s~
Hence deg (Tr t dxdy) and ~o
deg(Tr t dxdy)p C
P
= - (s-l)s §
an even integer~
is an eve~ integer.
...,
-49~11.
The a r i t h e m e t i c Let
algebrsic
F
genus
be s n o n - s i n g u l a r
curve on
F.
Let
surface
K = (w)
~nd P
an
irreducible
be any c a n o n i c a l
divisor,
and let Z be a divisor which is llnearly equivalent
to
and does no~ have
Z = ~ + (~)-(t)
~
ss a component.
For example,
K+ ~
i
where
t
is a unlformizlng
where
X
is 8n effective divisor and
on
.
If
F
/-7 h a s no s i n g u l a r
We r e s t a t e Prop.
11.1:
parameter of
If
77 .
K
Is a canonical
points,
then
divisor
an irreducible on
(1)
z~F§
(~)
p
F
and
Z
a b o v e ~s
c u r v e o n F, K a cycle
divisor
T r p Z = K.
T h e o r e m 1 C . 6 and t h e p ~ r s g r a p h T~ i s
Then TrrZ = K + X
a canonical
such that
and
is
not a component of
then the divisor
Trp Z
Tr r z = K ( p )
Z
of k(~)
+ s(~
has the form
)
I
where s(P)
K(D )
is
a c~nonicsl
divisor
of k(P)
(the "divisor ~f the singularities"
is an effective
snd
of
P
cycle each local P-component
of
~hich has even degree.
)
A prime divisor T of k ( D )
is a component of S( ~ ) if 8nd only if T is centered at
B singular p o i n t
s(p)
= o
of
7~
is and only if
.
In particular,
/7
h a s no s i n g u l a r
points. Let (Z.P) if
P
y (~)
- 2~ (p) is
be t h e + 2
an i r r e d u c i b l e
is
(effective)
s non-negative c u r v e on
77.
genus of
F
and
even integer. K
is
Then Furthermore,
a canonics1
divisor
-50on F, then (K. 7~) + (p 2) _ 2 7 / ( ~ ) integer.
We can therefore define (z.~)
Then
+ ( r 2)
(I)
p(P)
iS an integer,
(2)
P(P)
>- 7/" ( P )
())
P(T v) =
~- ( P )
-b 2 is v non-ne~ative oven p(~)
= 2p(r)
by the equation - 2.
and if and only if 7"
has no slngul=rlties.
We are thus led to the following Def. 11.2:
If
Z
iS any cycle on F, then the arlthmetlc genu.s
P ( z ) of Z is d e f i n e d by
~ot___e_e =
(a)
p(o) = l
(b)
p(K) = (Z 2) + l
(c)
p(-z)- I .
Prop. II.3 =
(z.z) + (z2) = 2p(z) - 2.
(a) p(zl+z 2) = p(z l) + p(z 2) + (zl.z 2) - l. (b) p(Z) is an integer. (c) If Z is an irreducible curve, i.e., a cycle, then p(Z) ~
p(Z) =
7F(Z).
prime
Furthermore
7/" (Z) if and only if Z has no Bin~ular
points. Proof= Hence
(a)
2p(Zl+Z2) = (K.Zl)+(LZ2)+(Z12)+2(Zz.Z2)+(Z22)+2.
P(Zl+Z 2) = P(Z l) + p(Z 2) + (Zz.Z 2) - z.
(c)
has a l r e a d y been proven0
-51(b)
By virtue of
If
negative of a curve.
that
p(Z)
is an integer.
a curve.
(-I,))=
P(-D) Assume
"~dJoint
is 7"
system of
By definition
Tr.pY
o2 t h e
is a curve we have already
Pence assume
Tr
using
z = p(7') + p C - a )
=
a non-singular I F I"),
of
the
(a),
I/~+KI
(the
177+Zl + 1.
TrrYIY e Ir§
I f ' q.K l =
divisor.
s y s t e m on
I-'
If
is non-singular,
I/'+I[ ! is
7v is
s y s t e m on ~
-1 ~ Tr r I P +xi ~- p C r ) - 1.
T i s ~ o t a componen'~
Trp
Hence
7" .
~anonical
w~ere 0 S 6(T' ) ~- p ( f ' ) .
and c o n s i d e r
We have
dim I KI § dim ~ r '
is a canonical
dimension
where
- ( r "~ - !
curve
~ and we have shown that if
canonical
Z = - T~
~hown
an integer. is
dim i r +KI
of Y ~
Z
Since p(O) = I, we have,
p(P+ Hence
(a) we can assume Z is a curve or the
is
a subsystem
non-singularj p(F
then
) - 1.
the Therefore
L e t dim ~ p i 1' +KI = P ( / ~ ) - l - O ( r )
Then s~.oe
dim I r + [ I = Pg + P ( r ' )
dim Izl = Pg - Z, we ha~,e - z - 8(r).
-52~12.
Normalization and complete systems. Let
V c Sn,
and let
(Yo' ""' Yn ) be strictly homogeneous
coordinates of a general point of R
is a graded ring;
V.
Let
R = k[VS = k[Yo~..~ + ..@
R = R o + R I + R 2 + ... + R q
set of homogeneous polynomial expressions of degree K = k(y)
(notes
Def. 12.11
K
is not
q.
is homogeneous if ~ = fq(y)/fs(y)
fq(y)
fs(y)
are forms of degree
Kq
K o = k(V)).
~
is
q-s
q
and
(and depends only on
be the set of homogeneous elements of
K
Without loss of generality we ce~ assume
then
~/yoq ~ K .
Since
Yo
Thus we see that
O
is the
k(V)).
~ ~ K
and
Rq
Finally, let
An element
The degree of Let
where
then
is transcendental ever
Kq ~ KoYoq
K~ ,
~Kq
s respectively. ~
).
of degree
Yo J O.
and
where
q
(hence
If ~ ~ Kq,
~Kq = Ko[Yo,I/Yo].
is a direct sum and hence
a graded ring. Let Yo
H'
be the integral closure of
is transcendental over
Ko,
R
in
K.
it follows that
is a graded ring and ~e can write
Rt =
~R"
Since
RI c
where
q_>O q IR' q
is a direct sum.
Rz I + ... + Rz h R' q
RT
zi
is a finite-dimensional vector space over
a point
U
in
ordinates of Let
A/k
and consider
Sm.
U.
and
V'
A'/k
L = k(A) = k(A t).
Let
A'
We say
exists a wluation
R' = R ' ~
and
Hence Kq
DI
and
q are homogeneous. k
for each
q.
Rt = Hence Let
(Uo, ..., urn) as the coordinates of
It is clear that they are strictly homogeneous co-
Let
respectively.
Ko[Yo].
is a finitely-generated R-module, so
where we can assume the
R'q = ku o + ... + kum
R C Ko[Yo]
v
B
of
be the locus of
U
in
Sm-
be biratio~lly equivalent varieties, and let and B L/k
B' and
be irreducible subvarieties of B'
A
and
are correspondin~ v~?ieties if there
such that
B
is the center of
v
on
A
and
-53B~
is the center of
A ~ B
v
on
At.
We say the birational transformation
has no fundamental varieties if to any irreducible sub-~ariety of
there corresponds only a finite number of irreducible subvarieties of Prop. 12~2:
The varieties
V
and
Vt
birational transformation rational transformation Proof:
Since the
ui/u j ~ K o
ui
for all
and all
from considering the elements combinations of the
u j)
k(V} = k(V' ).
V
Let
A I9
are birationally equivalent, the V: -~ V
V -* V I
is regular and the bihas no fundamental varieties.
are homogeneous elements of the s~ne degree, i
Thus
A
J~
yo%yoq-lYl , ..., y o q ' ~ n
that and
Vi = V - (V~Hi)
k ( U ) c K ~ ~ k(V).
Hence
(which are linear
yi/Yo ~ k(U), i = O, ..., n,
Vt
It is clear
and therefore
are birationally equivalent.
where
Hi
is given by
Yi = O, i ~ O, ..., n.
m
We have
yi q ~ J=~O cijuj' i = O, ..., n.
H lI. is given by k[V~]
(b)
t "'" V'n Vo'
is integrally dependent on cover
Since
we can write ui
(~
yoq This shows
ui/yoq
V'
and
ui/yoq
and
is integral over
kCV'o] = k[uo/yoq,...,Um/yqS~ k~oS.
u~ + al(~)u~'l + ... + a ~ ( y ) = O,
i.e.
)~
( V ' ~ HI ) where
-
Vt
is homogeneous of degree
of the ss~me de~ee,
=
We shall now show that
k[Vi];
k[VoJ = k[Yl/Yo, ..., yn/Yo S
To show (a) we must show R t,
V i'
J~=0 cijY'~ ~ O, i = O, ..., n.
(a)
We have
Let
aj(y)
+ al(Y>
yq
q,
where
ui
i~ in
ai(Y)E R.
we can assume all the ter~s a~e
is homogeneous of degree
(ui >@ -1 + ... + a~ (y)
U
is integral over
Since
k~o]
Jq.
Then we hav~
=0.
and proves (a).
It is sufficient to prove (b) for a rational point
p1 ' of
V ~.
There
is a zero-dimensional valuation be the center of for some
i.
v
on V.
v
with center
Since the
Vi
closed, and thus by (a) we have
k[V~] ~
on
V'.
Let
P
cover
V,
we have
Now
Rv
is integrally
R v D ~ ( V i / k ) D k[Vi].
Hence
P'
R v.
Hence
P' ~ V[
9
P~Vi
and
1
I
therefore the
Vi
It is clear that V~ * V
Vt
cover
.
k[V[] D k[Vi];
hence the birational transformation
is regular. To finish the proof we shall need the following well-known result: (I)
and
AI
If A
and
A'
are two noetherian rings such that
is integral over
A,
then given any prime ideal
there are only a finite number of prime ideals in to
~
in
Let
W'
valuation and
V
and assume
v
VI Let
W
in
A
W
and W'
Vo
A
which contract
carries an affine representative of
be a corresponding variety on of
k(V) = k(V')
VI respectively.
and so
in
A'
A.
~et W ~ W.
A'
~
AC
We have
V'.
Then there exists a
with the centers
W
and
R v ~'(~w(V/k)O k[Vo].
W:
Hence
on
V
R v D k[V o]
carries an affine representative of W' A = k[Vo] ,
and let
let
~t
A' = k[V~],
let ~
be the prime ideal of
are centers of
v,
we have
~'/%A
be the prime ideal of W'
in
- ~
A'.
Since both
9 Applying (1)
we have proved the proposition. ~Wrop. 12.38
Let kr be an irreducible subvariety of V 3 WI 9 e . e j
W'g
be the irreducible ~bvarieties of
correspond to and if i=l
W
and let
W.
Then
ovo
which
d l m W = dimW~, i = I, ..., g;
has an affine representative on Vo,
~w~(V'/k) = A~'
V'
where
A~
~ ~w(V~),
then and
Cot. 12.~,
The ~ o u p
of divisors of
divisors of Theorem 12.5;
If
q
v,q
Let
of
Then we can write
R.
coincides with the group of
(q _~ I).
is large, then
P~oof,
V
E = k[Yo, ..., yn ],
v'q
and let
is arit~aetically normal. RI
be the integral closure
R t= R o + R 1 + ... + Rq + ...
and
R' = q m
R' + R' + ... + R' + . . . . Let I = k[Uo, ..., um] where R' ~ F kui. o 1 q i-O (Note: I~_ R~.) We must show that I is integrally closed in its quotient field for large Since
I
noting that
q.
is a graded ring, we c~nwrite
IhCR
~q.
quotient field, then
Let
I'
be the integral closure ef
11 = I~ + ~
We ~ssert that
Iht = R'hq"
therefore over
I.
I = I o + II + ~
+ ... + ~
The elements of
+ ...
...
in its
I I C R' . h hq are integral over R and
R'hq
To show
where
I' = R' , we must show that h hq I. Let ~ be in ~ q .
in the quotient field of
I
+ ~+
R' hq
is contained
Then
= o and therefore
s~w
II = Since
~
~ 0'
+ R 2q t +
then
s i.
Let
9
+ R hq t +
.e,
I.
Hence we have to
9
is a finitely-generated R-module, we can write
R' = Rz I + ... + R z t degree
o C'< p D,P h(U>
is in the quotient field of
+ Rqt
Rt
P
nh(y )
where we can assume the
s = max (Sl, ..., st).
v1q
is arithmetically normal, i.e.,
For
q .> s
we clearly have
be any non-negative integer.
Then
+
are homogeneous and of
We shall show that if
RjRq_stZ t
q _> s,
I = II~
R'q = Rq-SlZl + "'" § Rq_stZt9
E tq*j = Rq+j_~l + """ + Rq+j_stZ t m R~q.SlZl + 9
zi
Let
J
-56It follows that we also have
Rt - RtRt. In particular, R' = r=t ~h q+J q J qh ~qJ ' IY = k + Rq + (Rrq)2 + ... + (Etq)h + . . . . Recalling that
and therefore
I = k[Uo, .. ., ~m]
R v = mZ kui, q i=O This proves the theorem.
I: = I.
where
The process of passing from of
V,
and Let
S
q, i.e,~ let
V ~q
V
we see that
to
vlq
!tC I
and hence
is called m normalization
is called a derived normal mo~eL
be z finite k-module of homogeneous elements of
S=
Kq = KoYoq
M = [6/~oI~
where
6S}~-
K o.
Ko = k(V).
Fix
~o ~S,
K
of degree
~o ~ O,
and
~e know there exists a smallest effective
cycle A such that (~ / ~ o) § A > O. ~ e set ~ ( ~ / ~o ) + A I ~ ~ S is called the linear syst'em of Let
Z~
that
be defined by
M
(~/~o)
and is denoted by = Z~
A = Z ~o . Hence we can write
we get
(~/~i)
L(S) =IZ~I Z~
= Z~
- Z~I.
- Z~,
-A.
LS(M).
Setting
( ~/~o ) =Z~
9
(See section %o
= ~o'
- Z~o
.
we see
Similarly
Thus we have a linear system
- (~/~t)f
.
The system
L(S)
has no fixed
components and is uniquely determined by thim condition and by the given k-module
3.
If degree
S = Rq, q,
then
i.e., if L(S)
h~permtvfaoes of order
S
for
L(Rq)
P~op. 12.6:
Let
and S
q.
If
f(y) / O,
Proof: Z~
and
Fix
f(Y) = O.
by the Zf
cal!~
We write
L r for L(Rq). q and S t be two finite k-modules of homogeneous q
are integral over
~o ~ S, 5o { O.
ZI
then we have a cycle
with the hy~ersurface
functions of degree S)
f(Yo, ""' Yn ) of
is called the linear system cut out on V
the intersection cycle of V Lq
is the set of forms
where
S~S
k[S],
For
~E
then S
t.
If the elements of
L(S)C L(S: ).
and
~t~ S:
we have cycles
such that
z~ - z~o = ( ~ / ~~
.nd Z'~, - Z'~o
= ( ~'/~~
Since
Z~o
is the smallest effective cycle
(~/~o) +A >O
and since
S E S v,
A with the property that
it follows that Z ~ o
~O
~
Hence, in order to prove the proposition, we must show only that (~'/~o) + Z ~ o -> O,
for all
~'
is homogeneous and integral over
~,r whei'e ~j(v) vo that
is in
in
S'.
+ ~ l ( v ) ,~, o - I
we can choose
~! = ~ !/vo
S = n~vj.
Since
~ ~
k[S], we can write + ...
is homogeneous of degree
S,
Let
J and
~o = Vo~
is integral over
+ a~, (v) = o
(l)
aj(v) C k[S]. Since
Dividing (1) by
v o~
k[Vl/Vo, ..~, Vm/Vo].
,
we s e e
Hence there
vi
exist polynomials AJ(~o), ,.,I' ~ ~et
r"
for all
occur in
where the degree of Aj + Al(vlvo).-.1' "*''l + . . .
Zvo with coefficient
J and so
v
(Aj(~)) >-J~. F
v
(~1') > - ~ P" propositlon~
o
Z(f)
Ci = E(Yi) , for
(~ > O).
T%Aen vr (vj/vo )
denote
Zf.
Prop. 12.6 shows that
Then Z(ylq ) = q Z ( y i ) ~ q
Let D
be a ~ cycle in
Let D - qCi = (~i),
[email protected] ~iYiq = ~jyjq
~i6k(V).
}o
~i'
and NV ~
S. S
Let
~
we have
are among the components of
is a minimal prime ideal in
S.
Ci where
Then
for all i,j. By definition
be the seb of all divisors
Lq~ Lv q
ILql ; then D = qCi (~i/~j) = ~i/~j = ~/yl 9
( ~o ) = D - qCo,
C o-
Consider S = k[Yl/Yo, ..., yn/Yo] , and let N closure of
.>'-~
> 0 which proves the o-
(yjq/yiq), and for a suitable choice of the
hence the poles of
(2)
Lq.
is integral over Rq~
i = O, ..., n.
i = O, ..., n.
+ A ~ ( v / v o) ., 0
Therefore, by (2) we see that
(~I/Vo) + Z v
determined by
Zet
such that
~ where ~Lq ~ denotes the complete linear system
T.qI ~
.=roof" Since R: q
< j,
-
and this shows
~heorem 12.8:
M
is
be the integral v
such that
Then ~ = ~ v~
Rv
S c Rv aud s~_uce
-58v( ~o ) -> 0
for all
~o yu~ ~ R'
v
N
~
for some u o.
00 = ~oYo q = ~iYi q. Iarge
in
it follows that Similarly
!
Now
RN+ q
for large
!
t
~ = ~h,
Cot. 12.8: V
and let
write ~
t = hq.
Therefore
D~ L' which will prove q
Then
~
CR' N+t'
~ ~ C R'tEN.
is arit~uetically normal if and only if
Let R ~
have t ~ s property,
and, if
This shows
k[~'t] and hence 03 is integral over R.
for all
in R.
N
Hence if we can show ~J is
we can conclude
is large, we have RN+ t ' = R~'t.
is integral over
of degree
N.
integral over R, i.e., OJ~ Rq, Let
Let
Then for large ~ we have ~ y i ~'q a R ~. For
( ~o ) = (uJ/yoq) = D - qC o.
the theorem.
and th~s
~ i Y i U i c R I for each i.
all mono1~ials in Yo' "" "' Yn
and so ~ c
~o m ~
Lq
is complete
q~
and R ~ be as before, and let ~
be the conductor of R'
is a homogeneous ideal and hence is graded. Therefore ~e can = J[o +
~ I + "'" + ~ h
+ "'" where
~o # @
if
~
is
the unit ideal. Pr~op~ 12.9: V
is normal if and only if ~
contains a power of the
irrelevant prime ideal (Yo, " "~, Yn )' Proof: Assume V
is normal.
Let the degree of &J since V
is normal,
Hence for some r k[y]
for some N.
be
Let GJ be any homogeneous element of R'.
q.
k[Vi]
we have
Then ~)/yiq
is integrally closed. Therefore ~o Yir~ k[y]
Since R'
where
~
some q.
k[y]. This shows that
(Yo' '"' Yn )h ~
is integral over This shows that
which shows that
and
b0/yiqE k~i],
&O(yo,. "" ,yn )NC .
has a finite basi~ there exists an integer ~
such that Rt(Yo, ..., y n ) h C Assume
is integral over k~i];
A i.
~
" Let Ai = k~i], Then
~yiq+hg R
~ yi q and so
(Yo' ""' Yn ) ~ and let
~k(V)
is integral over k[y] ~ y i q+h
I
is a form of
fo~
9 ~
-59degree
q+h. Therefore
~ E k[Vi] which proves
closed and so V
is normal.
Prop. 12.10:
is normal if and only if
V
Proof: Assume C
~
is normal.
for some
Therefore
h.
L'q = Zq
Assume say
V
q ~ h.
Lq
If and
q
is integrally
is complete for large
Then we have just seen that is large,
Lq
Lq
k ~ i]
R'= R q . h _. Rq h
(Yo' "'" Yn )h
and so RVq = Rq.
is complete.
is complete for large
q.
Therefore R'(Yo, "'" Yn )h c
Prep. 12.9 now shows that V
Then R' = Rq q R
and so
for large
is normal.
the equation of the hyperplane H
Then
be
k[Va] ,
V(~ ), the variety determined by ~
~
and ,
~[dh = (I) if and only if is contained in
H.
Under the
prime (primary) ideals go into prime (primary) ideals. dh be a variety (affine or projective). Let R be the co-
ordinate ring of V, and let Rt quotient field. Ps
Let ~ (V) and let W
= ~p(V/k) = ~r(V/k), Let
H(Y) = CoYo + ... + C n Y n = 0.
-~ ~
Let V
Let
Let
. We define
~ d h i S an ideal in
mapping
q,
(Yo''"~Yn)h ~ ~
Zet R = k[y], and let 07 be any homogeneous ideal in R.
Zet V a = V - V ~ H
q~
~p(V)
be the integral closure of R
denote the conductor of R' be the locus of
and let
~'
P
over
in its
in R. K.
Let
be the integral clo2ure of
denote the conductor of ~ '
in
~
.
The proof of the following proposition is obvious and we omit it. Prop. 12.11: Let V
be a projective variety, and let
i ~''" ~ ~t' (a) (b)
(V)= ~ o ~
where
T o is either (I) or is primary for ~ i
~
is a homogeneous primary ideal
Let V a be an affine representative of V
(Yo' "'" Yn )' (~i = ~ i and let
)"
P 6 V a.
Then
-6o-
(1)
t (Va) = ~ ~i, dh i=l t
~
~, dh
(2) /~p(v) o r] ~ i=l
(3)
dh = (~) if and o~ly iZ V( 9i ) i~ oon~i=od in V - V a .
~s
~4ith the conductcr of V.
~
It follows that
Cor. 12.12" Proof:
V
If
of V
V(~)o
P6V,
~
then V
is no.~mal at
P
we associate the subvariety if and only if
is normal at
if and only if
P
V
V(~ )
is normal.
if and only if
@ ' = ~.
P ~V(~).
This last condition
is equivalent to ~p(V) = (I) which in turn is equivalent to t P ~ ~)V(~i), i.e., to P ~ V ( ~ ) ( ~ i is defined as in Prop. 12.11). i=! Pro~. 12.13: If V is a h~persurface (affine or projective) of dimension r,
then ~
in R Proof"
(V)
is an unmixed ideal of dimension r-1 (both
and R').
It will be sufficient to deal with the affine case. Let
f(X I, ..., Xr+ I) = 0
be the equation of V.
with coefficients in the ground field f(X)
is monic in
where
Xr§1.
That is,
k
Xr+ I
Using a. linear transformation
if necessary, we can assume that is integral over
k[Xl, ..., x r]
R = k[Xl, ..., Xr+l]. Let ~
be a prime ideal in
~e .must show that ~ Since
R
: ~ = ~
and
R'
R (or R')
are integral over
is not principal.
dim ~
_
.
is a prime ideal of the same dimension as {~ A
such that
k[Xl, ..., xr] = A, ~
.
Since
~ #3 A
d~m ~
Therefore we can find polynomials
t~ A .< r-2, g, h g A /-)
which are irreducible and relatively prime.
C!e~ are in
~
~ ~_ ~ .
B(Xl, ..., Xr§
Let
: p.
oQ~R',
are in R
Let w ~ ~: ~ , Then where
then
wg(x) and
awg = A (Xl, ..., Xr§ A
and
B
and
wh(x) awh =
are of smaller degree ~
-61-
Xr+ 1
then the equation of integral dependence of Xr+ I
Since and
hA = gB, h
this relation must hold for the
are not associates, we have
~ = c(x)
is in R.
Let V
Therefore
be a variety, let
the integral closure of R k(V), RI
9
%hen
Then
K = Ko(Yo).
R'~ ,
Let
r Let
the
v
R
~t
~ 9 : ~
g
and
.
be its coordinate ring and let R t be K.
If we let
Ko
denote
be a minimal prime homogeneous ideal iu
v'
v'
of
K.
Since
has transcendence
the variety of the prime ideal dimension as ~ '.
v v to
K o.
Since
~t
is homogeneous~
is transcendental over the residue field of v
is a prime divisor of
v
and so ~
the residue field of
Therefore the residue field of
shows that
w g~
be the restriction of
v
since
k.
v'-residue of Yo
and hence
Finally,
is the valuation ring of a prime divisor
over
k[xi,...,Xr].
A/g = B/h = C(Xl, ..., x~+ I)
in its quotient field
t.d. K/k = r+l (r = dimk V), degree
X i.
over
v.
has transcendence degree r-1 over k(V).
The center of
~ = ~' ~ R
where
~
Hence the projective dimension of
v
k,
on V
is
has the same ~
is r-1 which
is a ~rime divisor of the .first kind. ~with respect to V.
Conversely, any prime divlsor of the . first kind with respect to V rise to a min~al homogeneous prime ideal in
R'.
gives
Thus we have a (l-l)
correspondence between prime divisors of the first kind with respect to V
and minimal homogeneous prime ideals in R t . Again let
let
~
~ I be a minimal homogeneous prime ideal in R t,
be the corresponding prime divisor.
and
Since ~ v is homogeneous,
co
we can write
~ ' =m=l ~* ~B tin where, of course, ~ m
the associated linear system, and let LS(~'m). = L'm " ~" ' i.e., LS(~Vm) Since
L' m
m
LS(R') = L' LIA
m"
It is easily seen that
is the set of ~--residues of
is complete, we know by Theorem %o7 that
q.
LS(~ ' ) is complete. m
-62If
~'
Let
is
~'-primary, then ~ ' = ~'(%))
for some
OT be an unmixed homogeneous ideal in R'
~ension r-l. ~hen we ~
~ite ~ o ~[(~-) ~
%2 . Hence
of projective
~ 2
where the ~ i ' are minimal homogeneous prime ideals in R t. Let q be the prime divisor corresponding to
L~- (~iux
~ i" !
Then
LS(6~m) =
§ ... § ~ r s ~ ~.d ~ ( ~ m ~ i, oo~p1~t~ ~o~ ~ n
In particular, if V is a hypersurface in St+l, then a prime (r-l)-dimensional ideal and If {~l(y), ..., ~h(y)} is a
LS(~m)
k basis for
~
~.
is
is complete for all m. ~m'
called subadJoint forms of the hypersurface V,
then the
~i(Y) are
~ d the hypermxrfaces
determined by ~i(Y) = 0 are called subadjoint hyoersurf~ces of V.
Hi
-| The Hilbert characteristic function and the arithmetic gen~s of a variety. Let I = ~
Im be a finitely generated graded module over the
polynomial ring k[Yo~...,yn] . ~(Ijm~
= dim k Im and call ~
We define the function
~
(Ijm) by
(I~m) the characterlstic function of
I. An important special case is the one in which I = k[Y]/O~ , where 0"{ is a homogeneous ideal.
In this case we shall also v~Ite
for The following theorem is due to D. Hilbert and J.P. Serre: ~(Ijm)
is a polynomial in m for large m: m
(IJm) = ~i(m) = .O(~) + al(r_l) for large m.
Here the constants
necessarily integers,
since ~
+ ... + a r
ao, al,...,a r are
(I~m) is an integral
valued f u n c t i o n . Furthermore, when I = k [ Y ] / ~
for di(m))~ it
c a n be s h o w n t h o r
r = dim V(~)~
then a o is the number of points in which V ( ~ ) complementary linear variety,
9 13,.1.:
then ideal Def. 13.2:
and
i f 0-(
(-1)r(ar-1) ~
If V is
is
called
the arithmetic
ar = r
(0)
g.enus o f t h e
pa(07).
i n Sn, t h e n b y t h e a r i t h m e t i c _ gegu..s
P a(V) o f V we mean t h e a r i t h m e t i c
o7 = I ( v ) .
prime,
i.e., a o is t h e order (or degree) of
and i s d e n o t e d b y a variety
is
is met by a
is a homogeneous ideal in k[Y] and i f
If ~
(m)
(in which case we write ~
genus of the ideal
-6~-
it c~-
i(v), ~,e =ay =rlte
L e t V be an i r r e d u c i b l e ~=
r-dimension~l
I(V) be t h e c o r r e s p o n d i n g k[Y]/6~
R-
~ll (v~m), Cv ~or variety
~
r
i n Sn, and l e t
homogeneous p r i m e i d e a l .
be t h e homogeneous c o o r d i n a t e
(v~=) = d i ~
)~ (C#jm),
Let
r i n g o~ V.
Then
= i + d i = "m' ~nd f o r l a r g e m n~
(vj~) = ao(rTM) + ai(r_i) + ... + ar_i(~) + a rLet R' be the integral
closure of R.
Then
~(~ (RZ;m) - dimkRtm
-- i + dim L m , and for large m ,
9
)~ (R, m~) -- CR, (m) = a
, ) + ~i(r,-l) + "'" + a'..
We shall show that rt = r and a o = act. conductor
Then
~
of R' in R where $(y) is homogeneous of degree h.
~(y)~
C ~§
and, o f c o = s e ,
~RCm) ~ ~R~(m) ~ ~ R ( ~ N ) , Therefore
Let $(y) be in the
~R
coefficients,
Rm ~ R='.
Hence
for h a fixed integer and for all large m.
and ~Rt have the same degree and the same leading i.e., r = re ~nd ao/r! = ao'/rl.
Hence
a O = ao'-
Let q be large enough so that Vr = v~q is arithmetically
nermal.
Then
Cv, (~) = r
= 1 + dim Lr.\q = ~
(mq), i.e., Cv, (m) = a o, q r ( ~ l +
~v~(O) = ~RI(O). models
~(V'~m)
Therefore,
(R';mq).
Hence
+ ar
Thus
...
for lar8~ q, all derived
VZq of V have the same arithmetic
normal
genus, namely
(-i)r(a~ - i). Let V be a normal variety,
and let D be an effective divlsorlai
cycle on V, i.e., D = m l ~ I + ... + m h / V h ,
m i > O for all i.
-65-
(m.,)
Let ~ i = I(#~i) in R = ~[V]. ve v l s h
~o~
to study
X(~,~)
De~. 13.3 t
r
~.
~o~
The a r S t h m e t i c
genus
(-1)r-l[r
consequently Proof=
p a(D)
iS
p ~(r
i.e.,
- I].
(D~=) = di~ ~
Prop. I~..4:
(~.
X (R/6"/j~), a,',d we writ~ X (Djm)
~ (6"/3m) =
~d
('~2)
Let I = R / ~ .
- di=(L=-D) for a l l m, and
~D(m) = dlm ~
Then Im = R m / ~ m
)d(Dj~) = dlm k ~m - d i ~
~
~.
- d i ~ ( ~ - D ) for large
and therefore
We know that d i ~
~==
l + dim
~.
Slnc~ LS(~/=) = L= - D, we h~ve d i % ~ Y = = l + di~(Lm-D) which proves Cor.
L_
the first part 8nd hence the proposition.
13.5:
i
J
I f D ~ D t w h e r e D and D! a r e b o t h ~D = ~ D t '
Proof|
If m is
and,
large,
in particular,
Lm-D i s
Lm - D =
D[=
effective,
then
P a(D) = p a(Dt)"
8 complete
]L m -
Lm -
system.
Hence
D'.
Prop. 13.6= If X and Y are effective divisorial cycles on ~ nonsingular Proof=
surface,
In view of Cor.
13.5,
t h e n ~X § ~Y = ~Xq-Y + ( X . Y ) . we may a s s u m e X and Y h a v e no
common c o m p o n e n t . L e t W1 and W2 b e s u b s p a c e s s p a c e W.
Then i t
is well
dim(Wl+W 2) + dim(W1/~W2). in R = k[Yoj..,3yn]. degree
of a finite-dimensional
known t h a t Now l e t
m i n A and B r e s p e c t i v e l y .
dim(A+B)m § dlm(A/~B)n.
Since
vector
dim W1 + dim W2 =
A and B be h o m o g e n e o u s
L e t Am, Bm d e n o t e
the
spaces
ideals
of forms of
Then dim Am + dim Bm = ~
~.
(A~m) + dim A m = dim Rm, e t c . j
-66we find
r247 + r
B= r
+ ~B"
L e t A = I(X) and B = I(Y).
Then I(X+Y)=
~X+Y"
~A = ~X' ~B = ~Y and ~A~B =
definition
sufficient to p r o v e that
By
A (~B. Hence i t
will
be
~A+B = (X.Y).
Let Pl,...,Pt be the common points of IX] and [Y], and let ~ i = I(Pi)" 7t
Since V(A+B) = [X]~[Y],
where
7i
is
~i-prlmary
we have A+B = 7o n ~ i ( ] . . .
f o r I <- i ~- t and
for the irrelevant prime ido-I (yo,...,yn). Vo
We have
D (yo,...,yn)h for some integer h, i.e., all forms in the Yi
of degree t Qi--
W~o is primary
~- h are i n
1~o"
Then
~ ~i" J=l
Hence
Let
v( 7 i + %) = r and so r q i , ~ i ~
o.
jWi Therefore,
again using the dimension argument,
CA4.B- ' ~ 7 1 + ' ' ' +
r
dim v ( ~ i ) =
we
o.
where the
t
dim
%'/%(~,~)
~ p rood ~p(~,~).
= u
.
Let
constants
since
= l(x,Y)Pj).
and let
o f X rand Y r e s p e c t i v e l y
are all
i
sh~ll show r
Let P be any of the 5 , equations
r
we h a v e
~ = O, ~ = O be local
a t P.
Thon i ( X , Y ~ P ) =
[~1,'",
~vt
Since the elements of
be s k - b a s l s of
C~(~,~)
are quotients
F(y)/G(y) of forms of the same degree where G(P) ~ O and F(y) e ~7 , we can choose are forms of degree
~i
== fi(Y)/Yo~
/u
.
To show
sufficient to show that
r
(m) =
~
~7
where the fi(y) = i(X,Y~P), it is
for large m.
This is an
obvious consequence of the following two simple assertions..
-67(I)
fl(y),...,f u (y) are linearly independent over k(mod ~
).
(2)
If m ~ / ~
form
, then the forms yo m-/a fj(y), J = I#..., ~
a basis of Rm ~ o d Cor. 13.7=
vT).
If X and Y are effective dlvlsorial cycles on a nonsingular surface~ then
p a(X+Y) = p a(X) + P a(Y) +
(x.Y) - l. In ~ I genus.
we gave a different definition of the arithmetic
Now we want to show that the two definitions are
equivalent.
If V is an irreducible curve, we shall denote by
(V) the effective genus of V. our
surface
F , we d e n o t e
by p(D)
the
arithmetic
We first prove the f o l l o w i n g
defined in ~ I . . P r o p . !3.-.8=
If D is a divisorial cycle on
Let V be an irreducible
curve
tv (=) = acre " P a(v) + l. and Proof z
p a(V) =
g e n u s o f D, a s I
in
Sn.
and l e t
Then p a(V) >- vr (v),
77" (V) if and only if V is non-slngular
We have ~V(m) = I + dim Lm, hence dim L m = aom - p a(V).
If Vt is a normalization of V, then dim L mt = aom - p a(V t). 9/ be the degree of the divisors in L 1 (hence of V).
Then the divisors in L mt
have degree
Roch theorem shows that dim L mt = m 7)
-~ m~
- 7/(V).
is the order The Riemann-
Therefore
!
= a O and
P a(V !) =
77"(V).
Since L m
C L m, we have
a consequence
of the obvious
p~ (v) >- Ps (v,) = rr (v). The l a s t equivalence
assertion
is
of the following
(a)
Pa(V) =
~
(b)
Lm = ~
for
(V). large
m.
statements-
Let
-68-
(c)
V
is norual.
(d)
V
is non-singular
(since V is a curve).
From this proposition ~nd frou ~ I
it follows that
ps(V) = p(V) for any irreducible nPnysingular curve on F.
From
Prop. ll.3(a) and ~bove Cor. 13.7 it follows that ps(D) = p(D) for any curv___~eD on F (i.e., any effective dlvisorial cycle D all couponents of which enter in D with coefficient I) such thst each irreducible component of D is non-singular.
We now can sketch
the proof of the general ProP. 13.9:
If D is 8n effective divisorial cycle on a nonsingular surface F~ then pa(D) = p(D).
First we assert without proof that the system L I 0f hyper plane sections of F contains curves all components of which are non-singu1~r c urYesL.
Pa(Cl) = p(Cl)"
This implies that if C I ~ L I, then
Thus
Cm
T m, then Pa(Cm) = p(Cm).
Let q be l~rge enough so that there exist effective cycles
E sstisfying
F_~cD ~ L q .
Then
ILm+DI =
ILm.Fq - E l .
We s h s l l
~Iso zssuue ~ithout proof ~ speci~l case of the theorem of Bertini~
Iff E i s an effectlve cYCle a n d m is l~rge~ then the
Systeu ILm-EI contalns non-singular., _ , curves.. _
Let Z be 8 cycle
in ILm+q-E I all of whose components are non-slngular curves with multSplicity one. such that C m + D =
Then Z s ILm+DI. Z.
Z~ it must hold for D.
Hence there exists C m e L m
Since the proposition holds for C m and
-69If
p r o p , . 13.1o,
D is
any effective
cycle
o n F, t h e n
~D(m) = (D.e~) - p(D) + 1 where C m is any cycle In Lm| Proo2t
Assume D = X+Y where X ~ O and Y ~ O.
~D(m) = ~X(m) + ~y(m) - (X.Y). X and Y, then
Then
If the proposition is true for
~D(m) = (X.em) - p(X) + 1 + (Y.C m) - p(Y) + I-(X.Y)
= (D.C=) - i p(x) + p(Y) + ~x.Y) - I ~ = (D.C=) - p ( x + Y ) ! + = (D.c=)
+ 1
1
- p(D) + I .
Thus ~e need only prove the propositlon for an irreducible curve D For 8n irreducible curve D~ ~D(m) = D2 is the order of D.
m - p(D) + I where
Let Lm be the system cut out on F by hyper-
surfaces of order m, and let ~ hypersurfaces of order m.
F, mnd let
~
be the system cut out on D by
Let (yo,...,ym) be 8 general point of
(yo*,...,ym*) be a general point of D.
R = k[y], and let
L*. = Ls(~).
R* = k[y*].
~ence
Lm* =
~b
We have L m = LS(Rm) and
Lm"
!Therefore,
Cm* is a divisor in L* m' then C* m = D.C m
Prop: 13.11:
Hence If D is
if
or some C m in L m.
32 is the degree of the divisors in L divisors in Lmo
Finally, let
~m
Sinoe
is the degree o2 the
9J m = (D.C~) ~ny effective
cycle,
then
p(-V) --(V2)_ p(D) + 2-- p(V-K) where K is a canonical cycle. Proofs
The first equality follows from 1 = p(O) - p(D + (-D)) - p(V) + p(-D) - (D 2) - 1 .
-70By d e f i n i t i o n
(see the beginning
o f ~11) we h a v e
-(D.z)--, (D2) - ~ ( D )
+
2.
The second equality now follows from p(D-K) = p(D) + p(-K) (D.K) - 1 and from p ( - K ) = CO r 9 1 3 . 1 2 r
-
I.
p(-D') = p(D) where D! ~. D + K.
Theoreu~ '13;,.13s
For any dlvisori~l integer
N(D)
cycle D of F there exists an
such that for oil m >- N(D) we have
dlmlD + cml -- pa(F) + pC-D-C m) - I. Proofs
Assume first D = O.
large m
dim L m = pa(F) + p(-r
dim L m = ~F(m) - i. all m. m~
In this case we must
- i.
We know that for large m
We ~ssert thst ~F(m) = Ps(F) + p(-C m) for
If m = O, we know #F(O) = 1 + pa(F).
dim Lm - dim(Lm-D ) = ~D(m)~
that CF(~) - ~F(m-1) = r induction on m.
show that for
Since~
for large
and since Lm_ 1 = Lm-C I, it follows
~r
assertion no~ follows by
Thus the theorem is true for D = O.
There exists an integer m such that Lm-D contains effective then dim
cycle
E ~
E,
C m + q - Z.
Then
D + Ca ~
CE(m+q) = (~..Cm+q) - p(E) + 1 .
by Car, 13.7.
dlmlD+cml
Let Z = D + Ca,
We hsve already observed
ICm+q-E{ = dim Lm+ q - ~E(m+q),
r
C m + q - Eo
C 2 = (re§
By Prop.
an
that
13.10
Hense
" P(%+q)
- p(-z) + 2
Therefore
= dlmlc~+q-~.l = dim ~ + q - ( c ~ )
+ pCcm+ q) + p(-z)- 2.
-71Since the theorem if
true
dim ICm§
f o r D = O, we s e e t h a t
= dim Lm§ q ~ Pc(F) + p(-Cm+ q) - I.
Thus
dim ID+cml = p~(F) + p(-Cm+ q) - (Cm2q) + P(Cm+ q) + P(-') - 3. Now a p p l y i n g P r o p .
1 3 . 1 1 we s e e t h a t
dim ID + Cml - Pa(F) + p(-Z) - 1 w h i c h proves the theorem.
Remark. of c y c l e s
The expression pe(F) + p(-Z) - I is e numerlcal character Z.
L e t us d e n o t e i t
by
~ (Z).
D + Cm~ where D i s e g i v e n c y c l e end m i s (m ~ N(D), w here N(D) d e p e n d s on D), dimension of
IZ I.
In e l l
cases,
I f Z i s of the form sufficiently
then
~ (Z) g i v e s t h e
? (Z) is s o m e t i m e s r e f e r r e d
to as the y i r , t u a l dlmension~ of IZl.
A simple calculation leads
to the following alternative expressions of
( z ) -- p a ( F ) + p ( - z )
- 1
-- (z 2) - pCz) + p ~ ( F ) + 1
-. ~ [ ( z 2) - ( e . z ) ]
large
+ p~ (F)
= p(z) - (z.z) + Pe(F) - I.
~ (Z)z
-72
-
The Riemann-Boch Theorem. Let F be s non-singular on F.
surface ~nd D a divisorial cycle
Let K be s canonic~l divisor on F.
We define i(D),
the
index of spec!alt~ ' O f D, to be
i ( D ) = 1 + dim II(-D I. In this section we sh~ll denote the geometric genus of F by
pg.
if pg-- o, then
nor doe~
IK-DI
if D
This shows that if pg = O ~nd D is effective,
is effective.
then i ( D ) =
IKI doe~ not exist,
o.
The Riemann-Roch theorem s~ys that dim IDI ~-pa(F) + pa(-D) - I - i(D). Def. 14.12
We define the deficiency of 8 cycle Z, denoted by 5(Z), by 8(Z) = pg + p(Z) - i - dim
I~Kl,
where p(Z) is the arithmetic genus of Z. The following is a res%ztement of some earlier results. Cot. I~.22
If E is an irreducible non-singul@r curve, 5(E) is the deficiency of the system
~(~) P r o p . !%.3~
Proofl D+E~.
= p(E)
then
TrE[E+K [, i.e.,
- 1 - dim TrE[E§
If C ~ D + E ,
where E is an irreducible non-singular
curve,
8(D) d 8(C) p r o v i d e d
then
Let C! = C+K, Dt = D+K and E! = E+K. Hence we h a v e
(D.E) > O. Then C~ ~
Dt+E =
-73-
dlm {c, { = di~ {D, { + b y Cot. 7.12. Z.
dim
T,E{C, I +
Z ~ C t and E Is not a component of
Fix Z such that
We con consider Z.E as m cycle on E, hence TrEICt I is a sub-
~ystem of the complete system IZ.EI on E. in {Z.E I is j = (D.E).
Since j
th~ result
di~
- 2 + J.
>
O by o s s u m p t i o n ,
14.1
yields
we h a v e
(Z.E)
~ 2p(E)
- 2.
the Riemsnn-Roch theorem for curves to E obtaining {Z.EI = p<E) - 2 +
Using
j. ~enc~
- i + J.
dim {C'I = Pg + p ( C )
dim I v ' l
dlm~E{C'{
( 1 ) we h o v e
dim IC'I ~ dim ID'I + p ( E ) Vef.
The degree of the divisor:
(Z.E) = (Ct.E) = (E'.E) +(D.E) = 2p(E) - 2 + J where
Thus we c~n v p p l y
p(E)
(1)
1
(2)
- 1 - 5 ( C ) and
= Pg + p
Hence 8(D) - 5 ( C ) ~ p ( D ) + p ( E ) + ( E . D ) p ( c ) = p ( D ) + p<E> + ( E . V ) - l ,
,e
have
- 1 - p(C).
5(V)
Since
- 8 ( C ) ~ O, ~ s a s s e r t e d .
We sh~ll now assume the following speci01 c o s o of Bertinits Theorem~ For ony m ~ I, there exist non-singular irreducible curves in
L•o Cor. 14.4:
I f Cm d e n o t e s
a cycle
i n Lm, t h e n
5(Ci)
~ a(Ci+l)
for
i = 1,2,3,... Proof~
If C m e Lm, we can find a Cm_ I e Lm_ 1 and an irreducible
non-slngul@r C 1 in L 1 such that C m ~ C m _
1 + C 1 and (Cm.C1) = m(C12).
Since (CI ~) > O, the result now follows from Prop. 14.3.
-74prop.. 14.5:
If m is large,
' : L e t Cm
Proof~
pg + p(Cm)
+ K.
1 - 8(Cm).
-
I = P(Cm) p ( - Cm)
t h e n 8(Cm) = pg - Pa w h e r e Ps = p ~ ( F ) . By d
initio,
of
8(%),
Since m is 10rge,
[see Cor.
13.12],
IC I =
Theorem 13.13 allows
dim IC~l = p~ + p(Cm) - 1.
The
result now f o l l o w s . Theorem 1 ~ . 6 :
If D is
any cycle
such that
(D.C1) > O, t h e n
5(D) ~ pg - Pa" Prooft
Consider
the fact
that
non-singular
ILm-DI w h e r e m i s l z r g e .
for
sufficiently
curves
On the other hand,
E.
large
L e t Ca =
m,
D+E.
We s h a l l
ILm-DI c o n t a i n s
Therefore
irreducible
Then (D.Cm) = m(D.C 1) > m~
(D.Cm) = (D 2) § (D.E) ~ m.
we ~ust h~ve (D.E) ~ O.
a s s u m e known
Hence if m is l~rge,
5(D) ~ 8(Cm) by Prop. 14.3.
The
theorem now follows from Prop. 14.5. Remark, D' :
By d e f i n i t i o n ,
IN-K.
5(D) = pg + p(D)
Hence T h e o r e m 1 4 . 6 s a y s t h a t
or p(D) + p~ - 1 ~ dim I D ' I .
- 1 - dim I D ' I , p(D)
- 1 - dim
S~nce p(D) = p ( - V ' ) ,
that p(-D t) + p~ - i ~ dim IDtl.
this
where
ID'I ~ -p. mesn~
Now i(D t) = I + dim [K-D' I =
I § dim I-DI, and thus i(D') = O since the intersection number of -D with C I is negative and since therefore
I-DI must be empty.
Thus
Hence Theorem 14.6 is
dim ID'I ~ p(-D') § Pa - I - i(D').
the Riemann-Roch ID§
where
theorem for the speci~l systems of the form
(D.C1) ~ O.
-75The rest of these lectures will be devoted to the proof of the theoreu of Riemann-Roch, fundamental
lemma
or rather to the proof of a
(Enriques-Severi-Zarlski)
of the Riemsnn-Roch
on which the proof
theorem is based.
Theore~ of Riemenn-Roch.
surface a n d D
If F is a non-slngular
is any dlvlsorial
cycle on Fj then
dlu IDI = Pa + p(-D) - i(D) + 5 - I
where
~oof~ -
Ij
8 E O.
By Cor. 7.12 ~e h~ve di~ IDI - d i ~ w h e r e Cm i s an i r r e d u c i b l e
not 8 component of D. asserts
that
dim
non-singul~r
- C m)
Let Z be a fixed cycle such that 8 component of Z.
is irreducible
and non-singular~
curws
dlu
p ( c m) =
canonical Let
Prop. K(Cm).
11.1
o f K.
m.
Hence
TrcmlD+Cml 8 ~- O.
the Riemann-Roch
- p(Cm) § d i m
_c IZ.C m
Since C m
theorem for
IKECm) - ~..cml + X,
the eenus oZ Cm. ~nd where K(C m) i s a
o f Cm.
K be a canonical
component
large
D + C m and C m is not
IZ.Cml - 8 where
IZ.Cml = (Z.Cm)
~ (cm),
divisor
Z ~
is
Then Theorem 13 9]3
- d - 2 for
TrcmlD+Cml.
This e11ows us to write d = dim
~.re
in Lm which
Then we can again view Z.C u ~s a cycle on C m-
As such it is 8n element of
yields
curve
Let d = dim Trcm_lD+C m I 9
IDI - Pa + p ( - D
- dim ~c~ID+C~I
ID~Cml
Let
Cm(1)
we c a n t a k e
Furthermore,
divisor
on F such
e Lm w h e r e
Cm i s
not
Cm i ~ n o t
Cm(1) ~ Cm.
K.C m + C m ( 1 ) . C u
since
that
as our
a couponent
Then b y
canonical of
either
cycle D or
-76Cm(1)
$
vle can take D+Cm(1) 8S our cycle Z.
Then K(Cm)-Z.C m
---
(K-D).C,1 , and dim IZ.Cml = (D.Cm) + (Cm2) - P(Cm) + dim I(K-D).CmI+I. Hence dim IDI = p~ + p(-D-Cm) - (D.Cm) - (Cm2) + P(Cm)-dim I (K-D).Cml +5
-3.
S i n c e p ( - D ) = p(-D-Cm+Cm) = p(-D-Cm) + p(Cr, ) - ~ . C m )
-
-
l,
we hove dim IDI = Ps + p(-D) - dim I(K-D).Cml
+ 8 - 2.
(*)
To complete our proof ~e need the following result which we shall refer to as the Fundamental Fundamental
Lemaa:
Lemma:
For any divisorial
cycle D there exists en
integer N(D) such that the Trcm[D[ if m ~ N(D).
More precisely8
cycle linearly equivalent
is complete
if D I is any
to D ~uch that C m
is not a component of D I, then Trcm[D I coincides with the complete
system
IDl.Cml
on C m. For the moment let us assume the Fundamental Lemm8 and complete the proof of the Riemann-Roch
theorem.
Lemm~, T r % I K - D I is c o m p l e t e .
dim
TrcmlK-DI=
dim IK-DI - dim
If m is ~ r g e j
Therefore re have
IK-D-Cml -
ICm+D-KI contains strictly p o s i t i v e cycles.
ond so
dim TrcmIK-D I = dim IK-DI.
l,
then s by the
dim I(K-D).Cml = If m iS l a r g e ,
Hence dim IK-D-Cml = -I,
S u b s t i t u t i n g t h i s i n (*) we hove
dim IDI = pa + p(-D) - dim IK-DI + 8 - 2.
By definition i(D) = dim IK-DI + i. dim IDI = Pa + p(-D)
then
Hence
- i(D) + 8 - I,
Q.E.D.
-77We now turn to the proof of the F.L.
Prop.
If
7t
the
true
F.L.
for
is
true
any cycle
for
(Fundamental Lemma).
a given
cycle
D, t h e n
it
is
D1 ~ D.
To prove this we assume for a moment the following lemma: Let E be an effective cycle on F, and let ID] be a given
Lamina
complete linear system. M = M(D,E)
There exists an integer
such that if V t ~ IV[ and if Dz.C m ~ E.C m
for some m ~ M, then D ! ~ E. Proof of 14.7:
Let Z e ]Dl.Cml
(we assume that C m is not a
component of D). is effective.
Since D 1 ~ D, we can write D = D +E where E 1 Hence D.C m ~ ~ + E.C m. We take m ~ max. I N ( D ) , M ( D , E ~ .
Since m ~ N(D), then Z + E.C m ~ Trcm[D I because the F.L. is true for D.
Hence there exists D (I) ~ [D[ such that
(a)
D(1).C m
is defined, and
(b)
n (I).% =
+ E.C m
Since Z is effectivej we have
D(1).Cm ~ E.C m.
Applying
we have D (I) = E+DI where DI ~ 0 (since m ~ M(D,E)). D = DI+E a n d D ( 1 )
~ [ D I , we s e e
D'.C m = Z.
Z r TrcmlDll
Thus
that
D' ~ V1 .
Lemma 14.8,
Since
H e n c e D! c [D1]
and so [DI.Cml = TrcmIDI[.
8nd
Hence
the F.L. holds for D I. Proof of 14.8| h.
Let
P
be e prime component of E, of multiplicity
Let s be the order of the curve
C 1 is a plane section. h](~2)~.
P
, i.e., s = ( P . C I) where
Let m be chosen so that ms >- (DT.p) +
We shall show that this implies D I >- h ~
.
If we assume
-78this for the moment,
then by applying
components
of E we o b t a i n
is proved.
Hence we must
this result
a lower bound
to the other
for m and the p r o p o s i t i o n
if ms >-
show that DI ~ h
(D, .P)
+
p2)[.
hl(
Let ~ O.
>
be the
We must
multiplicity
show
of
cycle3
and
Since
in Dr.
Assume
"2/ ~ h.
Then X is an e f f e c t i v e
~
/~
I/ < h. is
not
Dt i s
Let
effecttve
X = DT - ' ~
a component of
X.
By
our choice of m we see that
ms > ( X . N )
(l)
9
m
We c a n w r i t e Hence
D~;C m = E.C m + Z w h e r e
X.C m + ~ ~ . C m = h ~ . C m +
Z is
an effective
( E - h ~ ) . C m + Z.
cycle
o n Cm-
Therefore
(2)
x.c m >- (h- ~ ) V .C m . Since
L e t P c ~ / 9 Cm. x and y as equation
uniformizing
of
Cm s t
P.
( r ~ = i(X~Cm~P ) . A y + Bx ~
~--
parameters Let
X a n d Tv r e s p e c t i v e l y
=
Cm is a n o n - s i n g u l a r
at
By ( 2 ) and
multiplicity,
~ = O and p.
~ = Cy + Dx~
.
the
(X. ~ ) ]/ ~ h,
P such ~
<;- 0~ l
that
we c a n c h o o s e y -
= O be local
O is
a local
equations
of
i( ~,Cm~P ) and >- 4r~ .
We may w r i t e
w h e r e At B, C, a n d D a r e
By d e f i n i t i o n
of
in
intersection
we h a v e s
i(X, ~'73P) -'- dim k c~/O(~,~) where
Let
we h a v e
C~p(F/k) and B,D ~ ~
at
curve,
inequality
~ (P.Cm) Q.E.D.
follows
= ms.
Since
)= o-= i ( P , CmJP),
>- dim k ~ / , ~ ( y , from this
~
(~, ~)
6_~
contradicts
(y,x~ (1),
Therefore
we m u s t h a v e
3
-79In view of Prop. any collection given tion
14.7
it
L of divisorial
any divisorisl such thet
cycle
Di-Z is
We may say t h e n
that
is
sufficient
cycles
Z there
linearly
to prove
Di h a v i n g
exists
L is c o f i n a l
with
cycles (up to linear equivalence).
the
that
Di i n t h e
collec-
to an effective
cycle.
totality
dlvlsorie
of all
Now3 if D is any fixed divlsorla
cycle then the cycles D I = D+C I form such a collection L. paper,
for
the property
a cycle
equivalent
the F.L.
In our
"Complete linear systems on normal varieties and a generallza
tion of a lemma of Enriques-Severi"
(Ann. of Math., 1952) we have
taken for L the collection of cycles Ci, i = 1,2,...
(D = O).
In
these lectures we shall take as fixed cycle D a canonical cycle K. In other words, we shall prove that for each i = Ij2,... there exists sn integer m E N(i).
N(i) such that
is complete if
This is precisely the original formulation of the
fundamental lemma, due to Severl. Surfaces", p. 67). surfaces
TrCmlK+Cil
(See our monograph,
"Algebraic
Our present proof of the F.L. for algebraic
(which can be easily extended to varieties) makes use of
regular differentials of degree 2 and is based on the precise relationships between
(which will be established in the next section)
the geometric
of a surface different,
in ~
concepts
of an ad~oint
and s u c h a r i t h m e t i c
and c o m p l g m g n t a r y m o d u l e .
~nd s u b ~ d j o i n t
concepts As i n t h e
surface
as conductor, classical
proofs
of the Italian geometers, we shall also have to project our nonsingular surface F into a surface F o in ~ , to F.
biratlonally equivalent
However, we shall have no need of the proposition that the
projection can be made in such a fashion that F o has only "ordinary"
-80singularities.
All
we s h a l l
require
is
that
the
biration-al
oorres-
pondence between F and F o have no fundamental points on either surface~ in other words~
that F be a normalization of F o.
The
existence of a projection F ~ satisfying this condition is sn immediate consequence of the "normalization theorem" of Emmy Noether.
~15.
SubadJoint polyn0mials. Let S be an integrally closed noetherian doumln~ and let L
be its quotient field. extension of L.
Let K be a flnite~ separable~ algebralc
Firmlly~ let R be a ring such that S ( R C K
such
that
(i)
R is integral over S, and
(2)
K is the quotient field of R.
We define the complementary module for all
u e R I"
Clearly
~R/S
~R/S
=
~ z e KITrK/L
ZU
and
r
S
is a module over both S and R.
Furthermore it is a finitely generated module~ and since S is integrally we m e a n t h e
closed~ set
in R (since 1
we h a v e R~ ~ R / S
R c ~a/s.
~R/S
=
e,,C).
Assume R = S [ y ]
is integral over S.
where
y is
a primitive
element
of
K/L a n d y
Let f(Y) be the ~ o n i ~ minimal polynomial of y
over L (whence f(Y) ~ slY]).
~./s
By the different
= (1t~,(~)s[y] ~nd hence
It is known thst in this case
Y~/S = f'(y)S[y].
Assume S is c Dedeklnd domain~ and l~t K be es above. be the integral c~osure of S in K.
Since
K/L i s
separable,
Let
Rt
RT i s
-81also ~ Dedeklnd domain. <~i = X ~ R I / S 3
Let ~
and let ~
R/S'
l'
= ~
Ri/S" ~ = X ~ R / S
be the conductor of R t in R.
can be shown that
Rl~_~c~t
i.e., if
then
Prop. I~olZ
= •
and that if ~
Then it
is Invertlble,
Let K/k = k{V) where dim V = r, K/k is separably generated and k is an arbitrary ground field which is maximally algebraic in K. Let ~Xl~...,Xr~ and ~yl,...~yr~ be two separating transcendence bases of K/k. Let ~ be a prime divisor of K/k where is of the first
ProofI is
kind with respect
Let
Rx b e t h e i n t e g r a l
let
Ry
be t h e
integral
(~(YI'
''Yr )
We p r o e e e d by i n d u c t i o n
known ( s e e ,
for
instance,
closure
o n r~
"Algebraic
t o k~x~ ~nd k [ y ] . of k[x]
closure
since
for
of k[y]
i n K, and in
K.
r = I the result
Functions
o f One V a r i a b l e " ,
by C. Chevalley). We may assume ~
is t r i v i a l
on k * = k ( x 2 , . . . , X r ) .
also assume that each y i ( i = l s 2 ~ . . . , r )
-residue,
because
if, say, ~
if Yl hos an algebraic
has a trsnscendental
is trivial on k(x2,x3,...,xr) ~nd
~ - r e s i d u e s we can replace it by Yl § Y2
(which has a transcendent~l Ja c o b i a n .
We may
~-residue) without changing the
'
-82We know that t.d. K/k* = 1 and x I is a separating transcendental.
Clearly k*(Yl,...,yr)/k* is a separably generated extension
of transcendence degree one, and algebraic
extension.
k*(Yl,...,yr)/k*. K/k*.
K/k*(Yl,...,yr) is a separable
Let Yl be a s e p a r a t i n g
transcendental
of
Then Yl is also a separating transcendental of
Hence I Yl'X2,''"Xrl
is a sepBrsting transcendence basis
of K/k. We adopt the following notations~ (I)
k I = k(Yl).
(2)
Rx,y is the integral closure of k[Yl,X2J...,x r] in K.
(3) • x , y = ~uO Rx,y/k[Yl, X2,...,Xr]" F u r t h e r m o r e we know t h a t (4)
x 1 and Yl a r e b o t h s e p a r a t i n g
(5) [x2,...,x r ~ and ~y2,...,yr~ bases of K/k I. Derivations by D I.
transcendentals
o f K/k*, and
are separating transcendence
i n K/k* w i l l be denoted by D*, and t h o s e i n K/k 1
We have
*
d (Yl'X2'''''Xr)
Dx I Yl =
and
a (Xl,X2,...~xr )
~I(Ye,...,Yr )
=
~a l ( X 2 , . . . , X r )
(Yl, Y2, ---~Yr )
(Yl, x2,..-,Xr)
Hence
(1)
(Yl,Y2,...,Yr) (x 1 , x2,.
99 , x r )
I(Y2,..-,Yr )
(x2,...,xr)
"
Dx 1
Yl
Let M denote the multlpliosti%.e set k[x~,5,...,Xr]
- )0 ~ , l e t
Rx; M and Rx~ysM be t h e q u o t i e n t
rings,
to M, o f Rx
and Rx~y respectively, and let
~Ugx;M = RxgM" ~Qx
with respect
-83 -
~ x , y ~ M = Rx,y3M" ~Jtx,y" Then it is immediately seen that Rx~M is the integrsl closure of k*[x I] in K, Rx, y#M is the integral closure of k*[y I] in K and that x~M
RxsM/k* [xI ],
=
l ].
Now~ the divisor ~ K/k*.
is trivial on k*, henne is ~ prime divisor of
Furthermore, we hzve
vo~(Xl) >-O, ~ (yl) ~ O.
the case r = I, and observing that v~o (x~x,y,M) =
~(~x,y),
Hence, by
V~(~x,M)- ~(~x),
we f i n d t h a t
@
(2)
~(DxIYl) = ~ ( ~ , y )
- v08(~x).
Similarly, let N denote the multiplic~tive set k[y I] -~ 0~, let Rx,yjN and Ry, N be the quotient rings, with respect to N, of Rx, y and Ry respectively, and let ~ x,y~N = Rx, y#N" ~gx, y" X~y,N = RysN. X~y.
Then RxjysN is the integrsl closure of
kl[X2,...,Xr] in K, Ry3N is the integral closure of kl[Y2,...,y r] in K, and we have
x,y N --
Rx,YJH/kl
x ~ yIN - 2
Ry3N/kl[Y2,~
Our prime divisor ~
is also a [prime divisor of K/kl, and It is
of the first kind with respect to the rin~s
k l[x2,...,x r],
-84kl[Y2j...,yr].
Hence, by our i n d u c t i o n
that v~CJx, y;~)-- v ~ ( ~ , y ) ,
hypothesis,
>(~y,~)-
::
snd o b s e r v i n g
> (~),
we find that
c
-
The proposition now follows from (I), (2), and (3)Prop. !5.21
Let the notations 8rid 8ssumptlons be the same as in Prop. 15.1.
If
k
is algebralcslly closed, then
v~ Cdyldy~...dyr) - v~ (~y). Proofs
We csn choose
coordlnstes of
~
parameter o~ ~ -residues of
~ Xl,...,x r ~ to be 8 set of unlformlzing
, and we can 8ssume x I is z unlformIzing
(i.e., v~ (~I) = 1)o 'f ~2 '''"x-~ ~re the x2,...,Xr,
then the residue field
Z1
is a sepzreble 81gebrzic extension of k(~2,...,Xr). identify x i end xi for i = 2,...,r.
of We shall
Then, 8s before, we csn view
aS a prime divisor of K/k* where k* = k(x2,...,Xr). v~(xl) = 1 8rid since the residue field of ~ 81gebrsic over k*, the prime divisor
o v e r k*(x I). i.e., we hsve
Hence
0~
v~(~x)
coordinetes, we have
~
is separable
of K/k* is unrsmified
does not divide the different = O.
~(IX, M,
Since the x i are uniformizing ,~(YI' " ' ' ' Y r )
v~(dYl...dYr) = v~ L ~ [ x) , . . ~ . ~r)
V~(~gy) - V~(~X) where the last equslit7 follows from Prop. 1~.10.
Since
Hence >(dYl...dYr)= v. (~..).
=
-85T h e o r e m 15.3 |
Let V8 be an irreducible, r-dimensional effine hypersurface defined over an algebraically closed
ground field k.
Let f(Xl,...,X~§
= O be the
equation of V a.
Let R o = k[Xl~...,Xr.[.l] , let R
be the integral closure of R o in k(V) and l e t ~ be the conductor of R in R o.
Assume that
{Xl,...,x r ~ is a separating transcendence basis of k ( V ) / ~ and let differential.
~
= Adx I ....dx r be an r-fold
Then ~o is regular on V a if, and
!
only if, Aft+ I c ~ Proof!
Let d~
where fi: = ~ f/$xi"
stand for a prime divisor of k(V)/k, of the first
kind with respect
t o Va ( e q u i v a l e n t l y :
~
is of t h e f i r s t
The f o l l o w i n g
with respect to k[Xl,...,Xr] ).
assertions
kind ~re clearl:
equivalent~
(a)
~0
(b)
v0G(~) ~- 0 f o r
(c)
~6(A~x)
>- f o r
(d)
v~ (A ~ I
) >- v~ (~) ~or all
(e)
~(Af~+ l) ~ - ~ ( ~ ) f o r
is r e g u l a r
on Va . all
oq
.
e l l oco
.
an
r = 1 we see at once that
~
.
[By reduction to the case
vj~ t~ i0x ~ j \
--
.ft
r~l) ~" ))
is pure (r-l)-di~enslon~l, ~
Since where
~
I = ~ O e (~
runs through all the minimal prlme ideals in
R x = k[Xl, 9 .. .x r] and w h e r e
power.
h~
Thus ( e )
~b (v~(~))
is equivalent
to
is the ~,ymbolic prime
-86Theorem 15.3 regular
shows t h a t
on the affine
variety
any r-fold
differential
Va c a n be w r i t t e n
which is
in the form
r
(~/fr§
where ~ e ~
j and conversely.
Let V/k be ~ hypersurface in projective St,l, let V a be an sffine representative of V and let P be 8 poin~ of V.
Let
yo,Yl,...jyn be strictly homogeneous coordinates of the general point of V/k, let Xl,X23...~x n be the non-houogeneous coordinates of that general point (where we may ~ssume that x i = yi/Yo) and let
~
be the local ring of P on V/k.
An element w of the function field k(V) is said to be a subadjoint
in
function,
loc~!ly
8.t. P3 i f
W belongs
to the conductorj
~ , of the integral closure of ~
in k(V).
It follows that
a function which is subadJoint, locally et P, belongs to the local ring of P. An element w of k(V) is sald to be a subsdjoint function of the affine representative V a if w is locally subsdjoint at every point of V a. points of
Since the intersection of the local rings of all the
Va
is the coordinate ring k[Xl,X2~...,x n] of Va/k~ a n y
subadjoint function w of Va/k can be written as a polynomial expression of Xl,X2, ..OSXn, with coefficients in k. $(XI~X2,...,Xn)
in n indeterminates
A polynomial
Xi, with coefficients in k,
will be said to be s sub~djpint ' polynomial of Va/k if ~(Xl,X2~...,Xn)
is ~ subadjoint function of Va/k.
that an element w of k[Xl,X2,...,Xn]
It is clear
is a subadjoint function of
va/k if and only if w belongs to the conductor
(Va).
-87Finally, a homogeneous element w* of the coSrdinate ring k[yo,Yl,...,yn]
will be said to be a subadJolnt (homogeneous) '
function of the (projective)
conductor
~
hypervurface V if w* belongs to the
(V)~ and a homogeneous polynomial
in the n+l indeteruinates Yis with coefficients said to be a sub~dJoint, for~. pf
~(YoJY1,...~Yn+I) in k~ will be
V/k if ~(yo,Yl,...ayn)
is a
~ubadJoint f u n c t i o n of V/k. We consider an irredundant decomposition of (homogeneous) primary components and we denote by
~
(V) into
~a *
the
intersection of those pricmry components whose prime ideals do not contain Yo"
It is then immediate that if ~(XlJX2,...,~)
a polynomial of degree ~ ~(Xl~X2~...jXn)
is
with coeffloients in kj then
is a subadJoint function of Va/k if and only if
the homogeneous f u n c t i o n yn/Yo) belongs to
~a"
#(yo,Yl,...,yn)
= yJ@(yl/Yo, y2/Yo,...,
The f u n c t i o n ~(yo, Y l s . . . , y n )
i s not
necessarily itself a subajoint function of the (projective) hypersurface,
but since the primary components of
are not included in
(v) for ~ll ==fflclently large integers
It follows that every subadJolnt polynomial
of the affine variety
r
(V) which
~ a * all contain yoj it is clear that
yo~#(yo,yl,...,yn ) E ~ jA~ .
~
Va/h comes
~(XI,X2,...~Xn)
from a subadJoint form
of V/k by setting
Yo = I and Yi = Xi for i ~ O~
but it may be necessary to take for ~ a subadjolnt form of degree § will be
greater than the degree a
factor of ~(Y)).
~
of ~ (and in that case Yo~
Converselyj
it is obvious that if
-88~(Yo,YI,...,Yn)
is a subadjoint form of V/k, of degree g, then
~(1,Xl,X2,...,Xn)
is
a subajoint
polynomial
of Va/k
(of degree
g). = 0 be the i r r e d u c i b l e
Let F(Yo,YI,...,Yr+I)
homogeneous
equation of V/k, w h e r e F(Y) is ~ form of degree n, and let f(Xl,X2,...,Xr+ I) = F(I,XI, X2,...,Xr+I) irreducible equstion of V~/k. closed and that
So that f(X) = O iS the
We assume that k is algebreically is 8 separating trmnscendenee
Xl'X2~ "" ''Xr I
basis of k(V)/k. Theorem 15.4.
A necessary and sufficient condition that a differential cO = (A/ f~r+l(X))dXldX2"''dx r , be regular
on t h e
(projective)
A
k(V),
hypersurface
V/k
is that A be of the form ~(x), where ~(x) is a polynomial of degree ~ n-2-r such that
yn-2-r
(q/Yo, Y2/Yo'""Yr+I/Yo ) is
sub-
adjolnt form of V/k. Proof ~- We note the relations f
d X l d X 2 . . . d x r = (_I) r+i-I d X l d X 2 . . . d x i . . . d X r +
1 ,
i = 1,2,...,r, where
the
sign
~
a b o v e dx i s i g n i f i e s
factor dx i has to be deleted. st
once
that
if
that
this
differential
From these relations it follows
we make a c h a n g e o f v a r i a b l e s |
-89c i ~ k~
i = 1,2,...sr 3
z i ~ x i + CiXr+ I,
Zr+ I = Xr+ I s a~
if g(zlsz2,...,Zr,l)
= O is the irreducible
relation between
the zi, then
f !r + l
= g r!+ l ' d X l d X 2 "
dzldz2"''dZr
Hence in the new e~pression of th~ differential
''d:~r" ~
the adjoint
polynomial ~l(Z) which will occur is merely the transform of the polynomial $(x).
For a "non-special"
n I c i the term Zr+
will actually occur in g(z).
assume that the term f ( X ) = 0 o f Va .
value
of the r constants We thus may
~r+l occurs in the original equation
That means that
Yr+l = I does not belong to V. affine representative
the
point
H~nce,
Yo = Y1 =
if we denote by V i the
then the r+l a ffine varieties V i cover V.
Then ~
is regular on V if and only if it is regular
on each
V i.
~
= Yr = Os
V - V ~ H i of Vs where H i is the hyperplane
Yi = O (i = O~l,...sr),
For
"'"
to be regular on
is necessary and sufficient
it
that A be of the form $(x), where
$(x) is an adjoint polynomial adJoint polynomials $(X)
Vo
of V o (= Va).
Among the different
such that A = ~(x) we take one of
smallest possible degree h s and we let ~(Y) = Yoh$(Yl/Yo , Y2/Yo, ...,Yr+I/Yo).
We have to show that
the r afflne varieties VI,V2~...sV r and Yo n'2-r-h ~(Y) is a subadjoint
~
is regular on each of
if and only if h ~ n-2-r form of V/k.
-90~Te assume then that that
the
assumption
that
~
is regular on V and we shall show
k > n-O-r
leads
to
a contradiction.
Let~
~ = h-(n-2-r) > O.
then,
We set ~I = I / X l
...,r§
Then
= Yo/Yl 9 x i = x l / x l
Xl,X- 2,.. ., ~ + I <-
the Irredu~Ib1~ equatio~ o~ (% ~(X) = F , 1 , ~ 2 , . . . , X ~ + 1).
= Yi/Yl ' for i =
.,2,3,
) is a general point of vl /k, and
Vl/k is
g ( ~ l , ~ 2 , . . . , X ~ + l ) -- o, ~he~ __ _ F~ther~ore, ~ X l , X 2 , . . . , x r~ i~
obviously a separating transcendence ba~Is of k(V)/k.
A simple
calculation shows that
~l ~-r-2.*~ ~(~) (1)
~
=
d~id~ 2
d~ r
gr+l where ~(~) = ~(y)/yl h = $(X)~l h.
= ~
We have therefore
d ~ i d ~ 2 . . . ~ ~, ~, ~ o.
Sinc~ ~
i~ ~ u i ~
on v I
it follows, by Theorem 15.3, that there must exist 8 form As(Y), of degree s => O~ such that ~(x)/~l ~
r
= As(y) y J / y l ~
.
= As(y)/yl s.
Hence
This i~plies that y ~ r
divisible by Yo in k[y], ~or some integer z2 _-> O.
i~
If we repeat
the same reasoning for the afflne varieties V2,Va,...,Vr, we flnd similarly that y ~ ( y ) integer
~acaulay,
~
is divisible by Yo in k[y], for some
=> 0 and for i = O,l,2,...,r.
By a theorem of
the idesl (F(Y),Yo) in k[Y] is unmixed, of projective
dimension r-i =~ O, and the same is true therefore also of the principal ideal (yo) in k[y].
On the other hand, since the point
Yo = Y1 = "'" = Yr -- O, Yr+l = I, does not lie on V, the ideal
-91 -
(yo,Yl,...,yr)
is irrelevant
(of projective dimension -I).
It
follows that ~(y) must be divisible by Yo' in contradiction with
our c h o i c e o f ~ ( X ) . Ha~ing proved the i n e q u a l i t y
h ~ n-2-r,
we c a n now w r i t e
~(x) = ~(y'')/yon-2-r, where we now drop the condition that the degree of $(X) be minimal.
Thus, now ~(Y) is a form of degree
n-2-r.
~;
Then (I) shows that
is regular on V if and only if
~(y)/y~-2-r is a subadjoint function of the affine variety V i , for i = O,l,...~r. follows that
~
Since these affine varieties cover V, it
is regular on V if and only if ~(Y) is a sub-
adJoint form of V/k.
Cor. 1 5 . 5 t
This completes the proof of the theorem.
Xf Vl is
a
normalization of V, then
pg(v,) = 1 +
k
n_r_2(v),
i.e.,
pg(v') is e aal
to the number of linearly i n d e p e n d e n t subadJolnt forus ~(Y) of V os degree n-r-2. Note I t
If n < r+2, then pg(Vt) = O.
Note 2 |
The regularity of a differential
~
,
,
.
,
of course, imply that ~
, on
V,
does not,
is a regular differential of
the field
k(V)/k, unless V satisfies some further
conditions
(ass for instance, the condition that the
normalization of V be ~ non-singul~r v~riety).
special
adjoint
differential
polynomials
~(X) s u c h t h a t
The
the
~(x) f! dXldX2.~ r is regular for each r+l valuation whose center lies on the affine variety V a,
-92are called adJolnt polynomlals, and correspondingly one ~efines ~dJoint. fgrms of V/k.
The two concepts
(of subadjolnt and adjoint forms) coincide if the derived normal model of V/k is non-singular (in partlculsr~ if V has only so-called "ordinary s Ingula r it ie s" ). Let C/k be 8n irreducible curve in 5
closed.
where k is 81gebrsically
We say t h a t C i s a cOmplet 9 i ~ t e r s e c t i o ~
k[Yo~...~Y3] has
a
basis of two elements.
if
I(C)
in
A similar definition
holds in the affine case. Theorom 15.6t
Let C/k be an irreducible curve in affine 5 which is the intersection of two surfaces
f(x,Y,z) = o and g(x, L z ) =
o.
Let
Q be a
point of C and assume that one of the two surfaces has a simple point st Q.
Let (x,y,z)
be a general point of C/k~ and let, say, x be a separatinE transcendental of k(C)/k.
Then 8
necessary end sufficient condition for a dlfferenti81
~
of k(C)/k to be regular st Q is that
be of the f 9
~
~ (y,,) ~here ~
~ZQ(C).
dx
-93Proofs
The following conditions
a linear transformation
can all be realized by means of
and will therefore be assumed to be
satlsfied.6 (I)
Q is the origin.
(2)
The line X = Y = O meets C only at Qo
(3)
y is a primitive element of k(C)/k(x), function
and y is an integral
o f x.
(4)
Z is
integrally
dependent
on k [ x , y ] .
(5)
Z = O is the tangent plane of g(X,Y,Z) = O at Q where we are 8ssumlng 0 is a simple point of E(X,Y,Z) = O. Let ~(x,y) = 0 be the irreducible relation between x and y,
and let D be the plane curve $(X,Y) = O. (O,0) in the plane. = O~Q(C/k), shall show
~I
Since Q r C, we have P ~ D.
and let CY"I = O~p(D/k). = c~
of v on D is
over k[x,y],
(2).
P.
we s e e t h a t
of v on C because
v(x)
k[x,y,z]
C
Rv .
(~ "
We
=
8 lc
~nd z i s
Hence Q i s
(x,y).
~
, i.e., / ~
Consider
This ideal has only one zero, namely Q. is primary for the ideal
integral
the center
is integrally dependent on
~
Rv, i.e. s
> 0 and i n v i e w o f c o n d i t i o n
Let /;~ be the maximal ideal of
(f,gsX3Y)
~I G
such that
Since k[x~y] ~ ~
> 0 and v ( y )
This shows th.~t C*
We sh~ll show that ~
Clearly
Let
.
Let v be a valuation of K = k(C) the center
Let P be the point
= ~r
the ideal
Therefore
(X~Y,Z).
C9i" ysz).
(f,g,X,Y).
the ideal
We can write
-94-
g(x,Y,z) = Zh(Z) + ~(x,Y,z) + B(x, L z ) Y where h(O) r O.
Thus (~,g,X,Y)= (~,Zh(Z),X,Y).
we see that h(Z) ~ (X,Y,Z).
(~,~,x,Y)
= (x,Y,z).
Let u e ~Pi(X,y)
.
e h[x,y]
tha~
(x,y)
such that
= (x,y,z)
u - pi(x,y) e ~
L~l'
in k[x,y,z].
J- = 4"q" I' we have
The statement
"~is
i+l,
Since
~
is
(O~/~) contains a power of nwn.
~
large i, A~7 i+l c ~ I "
Since
and so
For each i ~ 1 there exists ~ polynomial
integrally d e p e n d e n t on Hence, for
Hence Z ~ (f,Zh(Z),X,Y)
It ~ollows
h(O) r o,
since
~
Therefore
Q = ~
6~=
~9 I.
p.
regular at Q" is equivalent
to each of
the statements is r e g u l a r
at P.
(a)
~
(b)
~o = Adx where A3~.# r Since
~
~o(O,O) ~ o.
=
~[,-Q =
~J~l' we can write z -
p" HI(z,y)/Ho(X,y ) where
Let h(X,Y,Z)- Z~o(X,Y) - Hl(X,Y).
we h~ve
h(X,Y,Z) = A(X,Y,z)~(X, L Z ) + B(X,Y,Z)g(X,Y,Z). Now either We have
n(O,O,O) r 0 or B(O,O,O) r O.
~ (h,g)/c~ (y,z)=
is a unit in
0 1"
A(x,y,z).
nssu~e A(O,O,O) r
9 (f,g)/~(Y,Z)
I t remains to prove t h a t
O.
and A ( x , y , z )
~ (h,g)/@(y,z)
and
!
g/y a r e associates in ~.~ . Locally at Q we have ~(X,Y) -- ah + bg.
Replace Z by H1/H o.
Then locally at P we have ~(X,Y) =-0 (mad g(X,Y, HI/He)). g(x,y,z) = O we have g(X, YtH1/Ho) =--O (mad ~(X.Y))
Since
locally St I).
-95Therefore b(x,y,z) is s unit i n @(X,Y) = a h + bE, we obtain
~9".
Differentiating
"~y ! = a ( x , . v,z)h; + b ( x , y,z )gyI
and
O = S(x,y,z)h'z + b(x'Y'z)gtz"
' = b ~z ~ @Y
Hence
unit in
~
.
~
[y,z)
Therefore
NOW
"
h'z
b(x,y,z)/hlz
=
which is
HO(X,y )
is a unit in ~
and so
!
$ y and
616.
~ (h,g)/e)(y,z)
e r e ~ssociat~s i n
Proof of the fundamental
lemma.
Let V be a normal v a r i e t y . divisorial
cycle
complete.
(Since V is normal,
L e t dim V = r ,
ZX
We s h a l l
such that,
and l e t
c~ .
for
all
determine
q ~ 1, Lq - ~
we k n o n Lq i s
(yo,...,ym)
a cert0in is
complete for
be ~ general
point
large
o f V.
q.)
By
the Noether normalization lemma we know there exist elements m
Zos...,Zr+ I ~here z i =
(1)
ciJYJ' cij r k, with the properties
~0
k(y) = k(,).
(2) k[y] = k[V] is integral over k[z]. Now k [ z ]
u~y b e r e g a r d e d
V* i n Sr+ 1 suitable
w h e r e V* i s
as the coordinate the projection
of a hypersurface
of V into
St+ 1
from a
Sm_r_ 2.
Let R = k[y] = k[V], integral
ring
closure
let
R* = k [ z ]
o f I1 ( h e n c e a l s o
o~ ~ in R*, i.e., ~
= k[V*],
of R*).
Let
and l e t /i
R be the
be t h e c o n d u c t o r
= Z (V*). ~ e n ~ = ~ 1 (~l)~.... ~-h(~h)
-96where the ~ i are minimal homogeneous prime ideals in R. is norm~l~ each ~ Let ~
~ i on V. Prop. 16.1:
Since V
i de%ermines a unique prime divisorial cycle =~/
I ~I
+'" "+
~h/7h"
If n is the order of V, then
Lq- ~
I~ ~ ~ q'~
= Lq- L3 = I Z d - ~
= lK+(q-n+r+2)Cl! 9
In particular, for q = n-r-2, Lq-g~ is the canonical system on V. Proof~ e ~q,
If ~ e %,
then
~
~- Z~ if, and only if, ~ e ~
Z~ ~ Lq s i n c e
then
L
q
Thus
q,,
Lq - ~
is complete, m e a n s Lq - g~ i s Since n is any regular
complete the order
differential
~o = _ r zn-r'~
T fr§
where
L
divisor of ~
n-r-2"
this
q ~- 1 .
o f V, i t ~
_
o
for
If
iS also
on V
is
zI
the order
Hence
o f V*.
of the form
zr
d ( ~o) ... d(~o)
Thus " ~, regular on V" implies that the
, (~), is in Lq -L~ = ~'Z~ - 4{} since for any prime
dlvlsorial cycle
]-7 on V for which z o ~ 0 we have
vr(~) -- v.,(r -
Vr(d(~ll'o)'" "d(~rl'o))-~r (fr+!l'
--r-2) + +
-
' vp (fr+l )
= vp (z~) - v,.,(~)
-- vrCz ~) - v.(~ ). If z o = 0 on
/-7 , we can find some z i such that z i ~ 0 on
/v
and repeat the argument above again obtaining vr(~ ) = vFCz~)-v p(z~ ).
-97~us
IKlCnn_r_ 2 - ~
systems are complete.
and hence I K I - - L n _ r _ The assertlon Lq -Z~
2 -~
, s l ~ c e both
= IK + (q-n+r+2)CII
is an obvious consequence. Let F be a non-singular (Yo'''"Yn)"
surface in Sn with general point
We project F onto a surface F* in 5
where we can
assume t h a t ~ (!)
(yosYl,y2,y3)
is a general point of F*.
(2)
F end F* arc birationally equivalent.
(3)
h[F] is integral over k[F*].
Let f*(Yo,Y1,Y2.,Y3) = 0 be t h e e q u a t i o n of F*. For each i n t e g e r
m -~ 1, v~e can choose h y p e r s u r f a c e s
Gm i n Sn
defined by a form g : ( Y o , YI,Y2, Y3) o f d e g r e e m so t h a t t h e f o l l o w i n g conditions
are satisfied.
(a)
The surface G: i n 5
defined by g:(Y) -- 0 IS non-singul~lr.
(b)
F/]G m is an irreducible non-slngular curve Cms and moreover C m is the complete intersection of F and Gm, i.e., the princlpal
ideal
(g~(y))
in k[F] differs
Cm i n k [ F ] e t most by en i r r e l e v a n t
(c)
If ( ~ o , ' " , ~ n ) Of
(d)
from the Ideel o f
p r i m a r y component.
i s the ~en~r~l p o i n t of C~, an~ C~ i s the l o c u s
( y o , Y l , y 2 , y3) in 5 '
then
(21)
Cm and Cm a r e b l r a t i o n a l l y
(3')
k[c=] i s i n t e g r a l over k i c k ] .
Cm i s e c o m p l e t e I n t e r ~ e c t i o n
equlv~lent.
of F
end the s u r f a c e G~.
Aotually it can be shown that (d) is implied by (e), (b), and (c).
-98Prop. 16.2: Proof=
If ~(yo,Yl,Y2, Y3) ~ ~
(F*), then ~(~o,Yi,~2,~3)
cL(Cm)
We first m~ke the following remmrk: Let C m be sn irreducible
curve on s non-singul~r
surface F.
In @enersl,
i~ ~ ~ C m mnd if
Q is simple for both C m ~nd F, then we c~n choose t mnd .nlt6r~zlng
ss
pmrmmeters o~ Q on F such thmt t = O is m locml
e q u m t i o n of C m at Q. and 8ssume
~
~
Le%
is regu!mr
~
be a differential
~t Q
Then Tr t ~ = Cm di~ferenti~l of degree one ~t Q becsuse ~ = A d
o~ degree two on F ~
is ~ regulmr
9
g = ~ d ~
where
sufficient
to consider
A r
(.~'Q(F/k)
~dt
~nd
~nd ~ r CY-Q(Cm/k ). ~ I t will be
the various mffine r e p r e s e n t m t i v e s
determined
by Yi = O, i = 0,1,2,3. * ~ = cm * - (C *m ~ He) where H o is Let F * = F* - (F*~ He) ~nd Cm,
the hyperpl~ne
Yo = O.
Let X i = yi/y O -~nd ~i = Yi/Yo ' i = 1,2,3. The
e q u a t i o n of F*s is f(XI,X2,~'~)=f*(I,XI,X2,E3)=O,~nd
intersection of ~(X) = 0 ~nd ~ ( X )
Cm, a is s complete
= 0 ~here g~(X) = ~ ( 1 , X ) . co=
are separating
transcendence
bsses o~ k ( X l , X 2 , x 3 ) / k
8nd k(Xl,~2,~3)/k
respectlvely. Then $(X) ~ ~ ~=
(~(x)/f
)dXldX2,
then
~
(F~) 8nd if
is regul~r on F a.
3 ~nd let
F~ be the affine r e p r e s e n t a t i v e
If Q is cny point of C m e ~t Q.
Let
to see that
~
= Tr
of F with respect
to Yo = O.
Fc3 then t = O is ~ loc~l equation of C m
~ . Then ~ is regul~r on . It is essy m t 8nd x I are u n i f o r m i z i n g coordlns%os of C m. Hence we
-99c~n write 8nd
~=
AdXldt
ab t ~'~ = g,x~
~t
and
+ ~,
~ ( ~~ ) "
This yields
r. ! g~ ~x2 ~3 f~
dXldt
dXldX 2 . . . . . .
dXldX 2.
f!
~3 Hence
=
r
-
8nd t h e r e f o r e
1.
<) {,x2, x 3 ) m
Since ~
.
is reguler on Cm, e, we h~ve $(~)
r~
(Cu,~)
which proves
t h e proposition. Let 8nd l e t kno~
~ ~
be t h e d i v i s o r i s 1 m be t h e d i v i s o r i ~ l
Tr C ~
(F*) i s an i ~ e ~ l
c y c l e on F d e f i n e d c y c l e on Cm d e f i n e d cont~ined
in
~
by ~ ( Y * ) ( d by ~
~ 0),
(C).
We
( C : ) where we n o t e
M
th2t
Trcu ~
IF*) ~ ( o ) .
homogeneous i d e a l
9~~
Tr C ~ m
The d i v i s o r
on Cm d e f i n e d
(F*) i s t h e i n t e r s e c t i o n
by t h e
cycle
~ .C a -
~'cm ~m"
Let Lq be the system of s~ctlons of F with hypersurfoces of order q ~nd let L"q,m be the corresponding system on Cu. q - I, Lq - ~
8nd
Lq,m -~u --
is complete for oll q ~ 1. complete for l~rge m.
~re complete.
For 811
Therefore Lg,m- ~ .Cm
We must sho~ that Trcm(L q- ~ ) is
-I00-
By Leuma 1 4 . 8 N(q)
such
in >- N ( q ) ,
that then
if
we k n o w t h a t D r
D >- / I
ILql 9
for
e~ch
E.C m = W
8nd
exists
a n d D.C m >- z~ .C m w h e r e Let
W - r
ITrcm(Lq-Zl
Then ~ + ~ . C m is of the form D.C m where becsuse ~ ~- O.
q there
Then If m ~- N(q),
8n integer
D >- 0
) l "~" ~q ,m - Zl .C m.
D c Lq ~nd so D.C m ~- z~ .C m
D = Z] + E where E ~- O~
T r c m ( L q - 4 ) is complete.
and
Hence
This cor~pletes the proof
of the F.L.
Offsetdruck: Julius Beltz, Weinheim/Bergstr
