Lecture Note
What is the Multipliity of a Base Point
-
University:
Amherst College -
Course:
MATH 271 | Linear Algebra Academic year:
2001
-
Views:
69
Pages:
22
Author:
Ralswiekx87
What is the
Multipli ity of
a Base Point? Outline
Bezout's Theorem
Degree of a Rational Surfa e
Serre's De nition
Hilbert-Samuel De nition
Combinatorial Computation
Algebrai Computation
Proof of Degree Formula
Other Stu (Time Permitting) Bezout's Theorem
Suppose that C; D P2 are urves
of degrees n; m with no ommon
omponents. Then
X
nm =
2 \
p
C
D
I (C; D)
p
where
~ ~gi
I (C; D) = dim OP2 =hf;
p
;p
~ ~g are lo al equations of
and f;
C; D near p. Degree of a Rational
Surfa e
' : P2
! P3 is de ned by a; b; ; d
of degree n, no ommon fa tors.
Z = fp j a(p) = = 0g is the
base point lo us.
S = '(P2 n Z ) is the image.
We assume dim S = 2.
Degree Formula:
n2 = deg S deg ' +
X m(p):
2
p
Z
m(p) is the \multipli ity" of p. Naive guess:
m(p) = dim OP2 =I = dim R =I ;
;p
p
p
where I = h~a; : : : ; d~i
p
Counterexample:
a = s2u + t3, b = t2u + s3,
= stu, d = s2u gives '
with basepoint p = (0; 0; 1)
and I = hs2; st; t2i.
p
The naive guess implies
m(p) = dim R =I = 3:
p
p
p However:
A Grobner basis al uluation
shows that the image surfa e
S P3 has degree 5.
Consequen e:
Sin e n = 3, the Degree
Formula gives
32 = 5 1 + m(p);
so that
m(p) = 4
in this ase. Serre's De nition
R lo al ring, m maximal ideal.
Assume R ontains k = R=m.
Then R-modules M; N with
dim M
R
k
N <1
have interse tion multipli ity
(M; N ) de ned by
X ( 1) dim Tor (M; N ):
0
i
k
i
R
i Example: Bezout Situation
Let R = OP2 , M = R=hf~i, N =
R=h~g i. Then
;p
~
0 !R !R !M !0
gives
! Tor1 (R; N ) ! Tor1 (M; N )
~
!N !N !M N !0
However, no ommon omponent
implies
~
N = R=h~g i ! N = R=h~g i
is one-to-one.
f
R
R
f
R
f Thus Tor (M; N ) = 0 for i > 0,
so that
R
i
(M; N ) = dim M
N
~ ~gi:
= dim R=hf;
R
Observations:
f;~ ~g form a regular sequen e
in the ring R.
hs2; st; t2i is not generated by
a regular sequen e. Hilbert-Samuel De nition
Let R, m and k = R=m be as
above and let M be a f.g. Rmodule. For ` 0, the Hilbert
polynomial implies that
e
+1
dim (M=m M ) = ` + : : :
d!
where d = dim R and e = e(M )
is the multipli ity of M .
d
`
k
Theorem: If dim R = 0, then
e(R) = dim R:
k Re nement:
Let I be an ideal with m M
IM for some s. Then ` 0 implies that
~
e
+1
dim (M=I M ) = ` + : : :
d!
~e = e(I; M ) is the multipli ity of
I in M .
s
d
`
k
Main Claim:
In the Degree Formula,
m(p) = e(I ; R );
for R = OP2 and I = h~a; : : : ; d~i.
p
p
;p
p
p Combinatorial
Computation
Let I k[x1; : : : ; x ℄h 1 ni = R
have nite odimension. Then:
The exponents of a monomial
basis of R=I give a nite set
E Z0.
Let C = Conv(Z0 n E ).
This gives the multipli ity:
n
x ;:::;x
n
n
Theorem:
e(I; R) = n! Vol (R0 n C )
n
n Example:
I = hs2; st; t2i k[s; t℄h i has
basis 1; s; t of R=I . Then
s;t
i
i
i
i
i
i
i
i
i
i
i i i
y i i
y yi
i
i
i
i
i
i
gives e(I; R) = 2 shaded area =
2 2 = 4.
Example:
I = hs2; t2i k[s; t℄h i has the
same multipli ity.
s;t Algebrai Computation
Assume m I R, where R =
regular lo al ring of dimension n.
Then:
If m J I R, then
s
s
e(J; R) e(I; R):
If in addition I J = I +1, then
`
`
e(J; R) = e(I; R):
J is a redu tion ideal of I . If I is generated by a regular
sequen e, then
e(I; R) = dim R=I
k
I is a omplete interse tion.
Two Important Fa ts:
I has a redu tion ideal whi h
is generated by a reg. seq.
The reg. seq. an be hosen
to be generi linear ombinations of the generators of I . Consequen e of First Fa t:
e(I; R) = min dim R=J;
k
where the minimum is taken over
all omplete interse tion ideals J
ontained in I .
This follows be ause
e(I; R) e(J; R) = dim R=J
k
holds for any CI ideal ontained
in I and be ause (by the rst
fa t) I has a CI redu tion ideal. Proof of Degree Formula
For ' : P2 ! P3 given by a; b; ; d,
we need to prove
n2 = deg S deg ' +
X e(I ; R )
2
p
p
p
Z
for R = OP2 and I = h~a; : : : ; d~i.
p
;p
p
Pi k a generi line ` P2. Then
deg S = #S \ `: We an assume:
` meets S at smooth points.
` meets S transversely.
' is etale above these points.
For oordinates x; y; z; w on P3,
let
` = H1 \ H2
for H1 : 1x + + 4w = 0 and
H2 : 1x + + 4w = 0. Then,
on P2, onsider the urves
C : f = 1a + + 4d = 0
D : g = 1 a + + 4 d = 0: Sin e Z is the basepoint lo us of
a; b; ; d, we have
C \ D = ' 1(S \ `) [ Z:
By Bezout's Theorem,
n2 = deg S deg '
X
~ ~gi:
+ dim OP2 =hf;
2
p
k
;p
Z
However, the se ond important
~ ~g generate a
fa t implies that f;
redu tion ideal for I . Thus
~ ~gi
e(I ; R ) = dim OP2 =hf;
and the theorem is proved!
p
p
p
k
;p Another Proof
We will use Fulton's Interse tion
Theory.
By p. 79, the Segre
lass of Z P2 is the 0- y le
s(Z; P2) =
X e(I ; R )[p℄:
2
p
p
p
Z
Then Prop. 4.4 implies
Z
deg S deg ' = P2 1(L)2
Z
(1 + 1(L))2 \ s(Z; P2)
Z
where L = OP2(n). Then we are
done sin e Z has dimension 0! The Rees Ring
This is the graded ring
R
Then set
1
M
+( ) =
f=
R
I
R
i=0
+( )
ii
I t
+( )
I =mR
I :
This is graded and nitely generated
over k = R=m. One an show that
dim R = dim R, whi h we denote n.
f
By graded Noether normalization, there
are generi ~s1; : : : ; ~sn 2 I =mI su h that
R is a f. g. module over k [~
s1 ; : : : ; ~
sn ℄.
f One
an show:
+( ) is nitely generated over
[1
n℄.
1
n are a regular sequen e.
R
I
R s ;:::;s
s ;:::;s
We may assume that the si are generi
linear ombinations of generators.
Now suppose that u1; : : : ; uN generate
R+ (I ) over R[s1 ; : : : ; sn ℄. Set
`
= max degree of
1
N:
u ;:::;u
Then one an easily show that
I
`+1 = hs1; : : : ; sniI `:
See Se tion 4.5 of Cohen-Ma aulay
Rings by Bruns and Herzog for details.
What is the Multipliity of a Base Point
of 22
Report
Tell us what’s wrong with it:
Thanks, got it!
We will moderate it soon!
Free up your schedule!
Our EduBirdie Experts Are Here for You 24/7! Just fill out a form and let us know how we can assist you.
Take 5 seconds to unlock
Enter your email below and get instant access to your document