Outlines
Elimination Theory, Commutative Algebra and
Applications
Laurent Busé
INRIA Sophia-Antipolis, France
March 2, 2005
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Outlines
Part I: Residual resultants in P2
Part II: Implicitization Problem
Part I: Solving polynomial systems with resultants
1
Overview of the classical approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
2
The residual resultant approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Outlines
Part I: Residual resultants in P2
Part II: Implicitization Problem
Part II: Implicitization problem for curves and surfaces
3
Implicitization of rational plane curves
Using resultants
Using moving lines
Using syzygies
4
Implicitization of rational surfaces
Using resultants
Using syzygies
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
Part I
Solving polynomial systems with resultants
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
The Macaulay resultant
Consider a system of 3 homogeneous polynomials in X0 , X1 , X2 :

P
α

 f0 = P|α|=d0 c0,α .X
α
f1 =
|α|=d1 c1,α .X
P

α
 f2 =
|α|=d2 c2,α .X
X α denotes a monomial X0α0 X1α1 X2α2 (αi ≥ 0).
The ci,α ’s are the parameters of the system with value in an
algebraically closed field k.
Theorem (Macaulay - 1902)
There exists an irreducible and homogeneous polynomial in
k[ci,α ], denoted Res(f0 , f1 , f2 ), satisfying:
Res(f0 , f1 , f2 ) = 0 ⇔ ∃x ∈ P2k such that f0 (x) = f1 (x) = f2 (x) = 0.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
The Macaulay resultant
Consider a system of 3 homogeneous polynomials in X0 , X1 , X2 :

P
α

 f0 = P|α|=d0 c0,α .X
α
f1 =
|α|=d1 c1,α .X
P

α
 f2 =
|α|=d2 c2,α .X
X α denotes a monomial X0α0 X1α1 X2α2 (αi ≥ 0).
The ci,α ’s are the parameters of the system with value in an
algebraically closed field k.
Theorem (Macaulay - 1902)
There exists an irreducible and homogeneous polynomial in
k[ci,α ], denoted Res(f0 , f1 , f2 ), satisfying:
Res(f0 , f1 , f2 ) = 0 ⇔ ∃x ∈ P2k such that f0 (x) = f1 (x) = f2 (x) = 0.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Computing the Macaulay resultant
Degrees: Res(f0 , f1 , f2 ) is homogeneous w.r.t. each fi :
degci,α =
d0 d1 d2
with i ∈ {0, 1, 2} fixed.
di
Computation: Let ν := d0 + d1 + d2 − 2, set A := k[ci,α ]
and consider the map (the first map of a Koszul complex)
3
M
A[X0 , X1 , X2 ]ν−di
→
A[X0 , X1 , X2 ]ν
i=1
(g0 , g1 , g2 ) 7→
g0 f0 + g1 f1 + g2 f2 .
It depends on the ci,α ’s and has maximal rank
Some maximal minors are the Macaulay matrices.
Their gcd (3 are sufficients) gives Res(f0 , f1 , f2 ).
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Computing the Macaulay resultant
Degrees: Res(f0 , f1 , f2 ) is homogeneous w.r.t. each fi :
degci,α =
d0 d1 d2
with i ∈ {0, 1, 2} fixed.
di
Computation: Let ν := d0 + d1 + d2 − 2, set A := k[ci,α ]
and consider the map (the first map of a Koszul complex)
3
M
A[X0 , X1 , X2 ]ν−di
→
A[X0 , X1 , X2 ]ν
i=1
(g0 , g1 , g2 ) 7→
g0 f0 + g1 f1 + g2 f2 .
It depends on the ci,α ’s and has maximal rank
Some maximal minors are the Macaulay matrices.
Their gcd (3 are sufficients) gives Res(f0 , f1 , f2 ).
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Computing the Macaulay resultant
Degrees: Res(f0 , f1 , f2 ) is homogeneous w.r.t. each fi :
degci,α =
d0 d1 d2
with i ∈ {0, 1, 2} fixed.
di
Computation: Let ν := d0 + d1 + d2 − 2, set A := k[ci,α ]
and consider the map (the first map of a Koszul complex)
3
M
A[X0 , X1 , X2 ]ν−di
→
A[X0 , X1 , X2 ]ν
i=1
(g0 , g1 , g2 ) 7→
g0 f0 + g1 f1 + g2 f2 .
It depends on the ci,α ’s and has maximal rank
Some maximal minors are the Macaulay matrices.
Their gcd (3 are sufficients) gives Res(f0 , f1 , f2 ).
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Recovering the Chow form
V (f1 , f2 )
Let f1 , f2 ∈ R = k[X0 , X1 , X2 ] be homogeneous polynomials defining isolated
points (a complete intersection) in P2 .
Introduce L(X) = c0 X0 + c1 X1 + c2 X2 .
L(X )
Proposition (Chow form)
Res(L, f1 , f2 ) =
Y
L(ξ)µ(ξ) ,
ξ∈V (f1 ,f2 )
where µ(ξ) denotes the multiplicity (or degree) of ξ.
⇒ An Absolute factorisation problem
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Recovering the Chow form
V (f1 , f2 )
Let f1 , f2 ∈ R = k[X0 , X1 , X2 ] be homogeneous polynomials defining isolated
points (a complete intersection) in P2 .
Introduce L(X) = c0 X0 + c1 X1 + c2 X2 .
L(X )
Proposition (Chow form)
Res(L, f1 , f2 ) =
Y
L(ξ)µ(ξ) ,
ξ∈V (f1 ,f2 )
where µ(ξ) denotes the multiplicity (or degree) of ξ.
⇒ An Absolute factorisation problem
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Recovering multiplication maps
The Macaulay matrix with
right degree in L :
E
..
 .
 L

..
.
F
..
.
And one may assume that there is
no solutions on X0 = 0:



f1 , f2 

..
.



Rν
E
F
A
B




C
D
X0 E
−−−
Rν \ X0 E
Proposition
The matrix of the multiplication by L in R/(f1 , f2 )|X0 =1 is given by:
ML = (A − BD−1 C)|X0 =1 .
⇒ The roots ξ are obtained through eigen-computations.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Recovering multiplication maps
The Macaulay matrix with
right degree in L :
E
..
 .
 L

..
.
F
..
.
And one may assume that there is
no solutions on X0 = 0:



f1 , f2 

..
.



Rν
E
F
A
B




C
D
X0 E
−−−
Rν \ X0 E
Proposition
The matrix of the multiplication by L in R/(f1 , f2 )|X0 =1 is given by:
ML = (A − BD−1 C)|X0 =1 .
⇒ The roots ξ are obtained through eigen-computations.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Recovering multiplication maps
The Macaulay matrix with
right degree in L :
E
..
 .
 L

..
.
F
..
.
And one may assume that there is
no solutions on X0 = 0:



f1 , f2 

..
.



Rν
E
F
A
B




C
D
X0 E
−−−
Rν \ X0 E
Proposition
The matrix of the multiplication by L in R/(f1 , f2 )|X0 =1 is given by:
ML = (A − BD−1 C)|X0 =1 .
⇒ The roots ξ are obtained through eigen-computations.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
The limitations
Problems:
If V (f1 , f2 ) is not finite then Res(L, f1 , f2 ) is identically zero.
If V (f1 , f2 ) has base points (roots which are independant of
the parameters ci,α ’s) then Res(L, f1 , f2 ) is also identically
zero.
Questions:
How to remove the non finite part of V (f1 , f2 )?
More generally, how to compute “a part” (even
zero-dimensional) of V (f1 , f2 )?
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
The limitations
Problems:
If V (f1 , f2 ) is not finite then Res(L, f1 , f2 ) is identically zero.
If V (f1 , f2 ) has base points (roots which are independant of
the parameters ci,α ’s) then Res(L, f1 , f2 ) is also identically
zero.
Questions:
How to remove the non finite part of V (f1 , f2 )?
More generally, how to compute “a part” (even
zero-dimensional) of V (f1 , f2 )?
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Three circles in the projective plane
Consider

 f0
f1

f2
the following system:
= c0,1 X02 + c0,2 X0 X1 + c0,3 X0 X2 + c0,4 (X12 + X22 )
= c1,1 X02 + c1,2 X0 X1 + c1,3 X0 X2 + c1,4 (X12 + X22 )
= c2,1 X02 + c2,2 X0 X1 + c2,3 X0 X2 + c2,4 (X12 + X22 )
Problem: When do they have a common point ?
f1
Two “base points”: (0 : 1 : ±i).
This implies Res(f0 , f1 , f2 ) ≡ 0.
f0
f2
P2
Refined question: When do they
have a common root which is not
(0 : 1 : ±i) ?
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Three circles in the projective plane
Consider

 f0
f1

f2
the following system:
= c0,1 X02 + c0,2 X0 X1 + c0,3 X0 X2 + c0,4 (X12 + X22 )
= c1,1 X02 + c1,2 X0 X1 + c1,3 X0 X2 + c1,4 (X12 + X22 )
= c2,1 X02 + c2,2 X0 X1 + c2,3 X0 X2 + c2,4 (X12 + X22 )
Problem: When do they have a common point ?
f1
Two “base points”: (0 : 1 : ±i).
This implies Res(f0 , f1 , f2 ) ≡ 0.
f0
f2
P2
Refined question: When do they
have a common root which is not
(0 : 1 : ±i) ?
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The Macaulay resultant
Solving zero-dimensional polynomial system
Limitations
Three circles in the projective plane
Consider

 f0
f1

f2
the following system:
= c0,1 X02 + c0,2 X0 X1 + c0,3 X0 X2 + c0,4 (X12 + X22 )
= c1,1 X02 + c1,2 X0 X1 + c1,3 X0 X2 + c1,4 (X12 + X22 )
= c2,1 X02 + c2,2 X0 X1 + c2,3 X0 X2 + c2,4 (X12 + X22 )
Problem: When do they have a common point ?
f1
Two “base points”: (0 : 1 : ±i).
This implies Res(f0 , f1 , f2 ) ≡ 0.
f0
f2
P2
Refined question: When do they
have a common root which is not
(0 : 1 : ±i) ?
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
The setting
We keep the same notation; R = k[X0 , X1 , X2 ].
Let g1 (X), . . . , gm (X) be homogeneous polynomials in R of
positive degree k1 , . . . , km , respectively.
Let d0 , d1 , d2 be integers such that di ≥ kj for all (i, j).
One considers the system :

Pm
 f0 = Pi=1 hi,0 (X) gi (X) total degree d0
f1 = Pm
total degree d1
i=1 hi,1 (X) gi (X)

m
f2 =
total degree d2
i=1 hi,2 (X) gi (X)
P
α
where hi,j (X ) = |α|=dj −ki ci,j
α X .
Main fact
The Polynomials fi are generic of degree di in the ideal
G = (g1 , . . . , gm ).
Rk : The case G = (1) corresponds to the previous setting.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
The setting
We keep the same notation; R = k[X0 , X1 , X2 ].
Let g1 (X), . . . , gm (X) be homogeneous polynomials in R of
positive degree k1 , . . . , km , respectively.
Let d0 , d1 , d2 be integers such that di ≥ kj for all (i, j).
One considers the system :

Pm
 f0 = Pi=1 hi,0 (X) gi (X) total degree d0
f1 = Pm
total degree d1
i=1 hi,1 (X) gi (X)

m
f2 =
total degree d2
i=1 hi,2 (X) gi (X)
P
α
where hi,j (X ) = |α|=dj −ki ci,j
α X .
Main fact
The Polynomials fi are generic of degree di in the ideal
G = (g1 , . . . , gm ).
Rk : The case G = (1) corresponds to the previous setting.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
The setting
We keep the same notation; R = k[X0 , X1 , X2 ].
Let g1 (X), . . . , gm (X) be homogeneous polynomials in R of
positive degree k1 , . . . , km , respectively.
Let d0 , d1 , d2 be integers such that di ≥ kj for all (i, j).
One considers the system :

Pm
 f0 = Pi=1 hi,0 (X) gi (X) total degree d0
f1 = Pm
total degree d1
i=1 hi,1 (X) gi (X)

m
f2 =
total degree d2
i=1 hi,2 (X) gi (X)
P
α
where hi,j (X ) = |α|=dj −ki ci,j
α X .
Main fact
The Polynomials fi are generic of degree di in the ideal
G = (g1 , . . . , gm ).
Rk : The case G = (1) corresponds to the previous setting.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
The setting
We keep the same notation; R = k[X0 , X1 , X2 ].
Let g1 (X), . . . , gm (X) be homogeneous polynomials in R of
positive degree k1 , . . . , km , respectively.
Let d0 , d1 , d2 be integers such that di ≥ kj for all (i, j).
One considers the system :

Pm
 f0 = Pi=1 hi,0 (X) gi (X) total degree d0
f1 = Pm
total degree d1
i=1 hi,1 (X) gi (X)

m
f2 =
total degree d2
i=1 hi,2 (X) gi (X)
P
α
where hi,j (X ) = |α|=dj −ki ci,j
α X .
Main fact
The Polynomials fi are generic of degree di in the ideal
G = (g1 , . . . , gm ).
Rk : The case G = (1) corresponds to the previous setting.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Existence
Notations :
F := (f0 , f1 , f2 ) ⊂ A[X0 , X1 , X2 ] with A = k[ci,j
α ],
(F : G ) = {f ∈ A[X0 , X1 , X2 ] such that fG ⊂ F }.
Theorem
If G is a local complete intersection and max(di ) ≥ min(kj ) + 1
then there exists an irreducible and homogeneous polynomial in A,
denoted Res(f0 , f1 , f2 ), which satisfies: for any ci,j
α →k
Res(f0 , f1 , f2 ) = 0 ⇔ V (F : G ) 6= ∅.
⇔ the fi vanish on a point of Pn “outside” the locus V (G ).
Rk: Macaulay resultants ⇔ G = R:
V (F : R) = V (F ) 6= ∅ ⇔ ∃x ∈ Pn : f0 (x) = f1 (x) = f2 (x) = 0.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Existence
Notations :
F := (f0 , f1 , f2 ) ⊂ A[X0 , X1 , X2 ] with A = k[ci,j
α ],
(F : G ) = {f ∈ A[X0 , X1 , X2 ] such that fG ⊂ F }.
Theorem
If G is a local complete intersection and max(di ) ≥ min(kj ) + 1
then there exists an irreducible and homogeneous polynomial in A,
denoted Res(f0 , f1 , f2 ), which satisfies: for any ci,j
α →k
Res(f0 , f1 , f2 ) = 0 ⇔ V (F : G ) 6= ∅.
⇔ the fi vanish on a point of Pn “outside” the locus V (G ).
Rk: Macaulay resultants ⇔ G = R:
V (F : R) = V (F ) 6= ∅ ⇔ ∃x ∈ Pn : f0 (x) = f1 (x) = f2 (x) = 0.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Existence
Notations :
F := (f0 , f1 , f2 ) ⊂ A[X0 , X1 , X2 ] with A = k[ci,j
α ],
(F : G ) = {f ∈ A[X0 , X1 , X2 ] such that fG ⊂ F }.
Theorem
If G is a local complete intersection and max(di ) ≥ min(kj ) + 1
then there exists an irreducible and homogeneous polynomial in A,
denoted Res(f0 , f1 , f2 ), which satisfies: for any ci,j
α →k
Res(f0 , f1 , f2 ) = 0 ⇔ V (F : G ) 6= ∅.
⇔ the fi vanish on a point of Pn “outside” the locus V (G ).
Rk: Macaulay resultants ⇔ G = R:
V (F : R) = V (F ) 6= ∅ ⇔ ∃x ∈ Pn : f0 (x) = f1 (x) = f2 (x) = 0.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Computing residual resultants (1)
Assume that G is l.c.i. and moreover a codim 2 ACM ideal.
This implies that G is completely determined by its first syzygies :
0→
m−1
M
φ
R(−li ) −
→
i=1
m
M
(g1 ,...,gm )
R(−ki ) −−−−−−→ G → 0.
i=1

p1,1 . . . p1,m−1


..
L := Mat(φ) =  ...
.
.
pm,1 . . . pm,m−1

Proposition (multi-degree)
The degree is completely explicit in the di ’s, kj ’s and lt ’s:
d0 d1 d2
degfi (Res(f0 , f1 , f2 )) =
−
di
Laurent Busé
Pn−1
2
i=1 lj
−
2
Pn
2
i=1 kj
.
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Computing residual resultants (1)
Assume that G is l.c.i. and moreover a codim 2 ACM ideal.
This implies that G is completely determined by its first syzygies :
0→
m−1
M
φ
R(−li ) −
→
i=1
m
M
(g1 ,...,gm )
R(−ki ) −−−−−−→ G → 0.
i=1

p1,1 . . . p1,m−1


..
L := Mat(φ) =  ...
.
.
pm,1 . . . pm,m−1

Proposition (multi-degree)
The degree is completely explicit in the di ’s, kj ’s and lt ’s:
d0 d1 d2
degfi (Res(f0 , f1 , f2 )) =
−
di
Laurent Busé
Pn−1
2
i=1 lj
−
2
Pn
2
i=1 kj
.
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Computing residual resultants (2)
Construct the m × (m + 2) matrix gluing L and the hi,j ’s:

p1,1

L|H :=  ...
pm,1
...
p1,m−1
..
.
h1,0
..
.
h1,1
..
.
. . . pm,m−1
hm,0
hm,1

h1,2
..  ,
. 
hm,2
Set ν := d0 + d1 + d2 − 2 − 2 min kj and let I ⊂ Nm be the
set indexing the m × m minors of L|H. Construct the matrix
M of the map (first map of an Eagon-Northcott complex):
M
A[X0 , X1 , X2 ]ν−deg(∆s )
→ A[X0 , X1 , X2 ]ν
s∈I
(. . . , bs , . . .) 7→
X
bs ∆s
s∈I
Proposition
This matrix “represents” Res(f0 , f1 , f2 ).
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Computing residual resultants (2)
Construct the m × (m + 2) matrix gluing L and the hi,j ’s:

p1,1

L|H :=  ...
pm,1
...
p1,m−1
..
.
h1,0
..
.
h1,1
..
.
. . . pm,m−1
hm,0
hm,1

h1,2
..  ,
. 
hm,2
Set ν := d0 + d1 + d2 − 2 − 2 min kj and let I ⊂ Nm be the
set indexing the m × m minors of L|H. Construct the matrix
M of the map (first map of an Eagon-Northcott complex):
M
A[X0 , X1 , X2 ]ν−deg(∆s )
→ A[X0 , X1 , X2 ]ν
s∈I
(. . . , bs , . . .) 7→
X
bs ∆s
s∈I
Proposition
This matrix “represents” Res(f0 , f1 , f2 ).
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Computing residual resultants (2)
Construct the m × (m + 2) matrix gluing L and the hi,j ’s:

p1,1

L|H :=  ...
pm,1
...
p1,m−1
..
.
h1,0
..
.
h1,1
..
.
. . . pm,m−1
hm,0
hm,1

h1,2
..  ,
. 
hm,2
Set ν := d0 + d1 + d2 − 2 − 2 min kj and let I ⊂ Nm be the
set indexing the m × m minors of L|H. Construct the matrix
M of the map (first map of an Eagon-Northcott complex):
M
A[X0 , X1 , X2 ]ν−deg(∆s )
→ A[X0 , X1 , X2 ]ν
s∈I
(. . . , bs , . . .) 7→
X
bs ∆s
s∈I
Proposition
This matrix “represents” Res(f0 , f1 , f2 ).
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Recovering the Chow form
Let f1 , f2 ∈ R = k[X0 , X1 , X2 ] be homogeneous polynomials defining some isolated points Z and V (G) (l.c.i. codim
2 ACM) in P2 .
V (f1 , f2 )
V (G )
Introduce L(X) = c0 X0 + c1 X1 + c2 X2
and choose f vanishing on V (G ) and not
on Z.
L(X )
Proposition (Chow form)
Res(Lf , f1 , f2 ) =
Y
L(ξ)µ(ξ) ,
ξ∈V (F :G )
where µ(ξ) denotes the multiplicity (or degree) of ξ.
⇒ An absolute factorisation problem
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Recovering the Chow form
Let f1 , f2 ∈ R = k[X0 , X1 , X2 ] be homogeneous polynomials defining some isolated points Z and V (G) (l.c.i. codim
2 ACM) in P2 .
V (f1 , f2 )
V (G )
Introduce L(X) = c0 X0 + c1 X1 + c2 X2
and choose f vanishing on V (G ) and not
on Z.
L(X )
Proposition (Chow form)
Res(Lf , f1 , f2 ) =
Y
L(ξ)µ(ξ) ,
ξ∈V (F :G )
where µ(ξ) denotes the multiplicity (or degree) of ξ.
⇒ An absolute factorisation problem
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Recovering multiplication maps
A Macaulay type matrix
with right degree in L :
E
..
 .
 L

..
.
F
..
.
And one may assume that there is
no residual root on X0 = 0:



f1 , f2 

..
.



Rν
E
F
A
B




C
D
X0 E
−−−
Rν \ X0 E
Proposition
The matrix of the multiplication by L in R/(F : G )|X0 =1 is given by:
ML = (A − BD−1 C)|X0 =1 .
⇒ The roots ξ are obtained through eigen-computations.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Recovering multiplication maps
A Macaulay type matrix
with right degree in L :
E
..
 .
 L

..
.
F
..
.
And one may assume that there is
no residual root on X0 = 0:



f1 , f2 

..
.



Rν
E
F
A
B




C
D
X0 E
−−−
Rν \ X0 E
Proposition
The matrix of the multiplication by L in R/(F : G )|X0 =1 is given by:
ML = (A − BD−1 C)|X0 =1 .
⇒ The roots ξ are obtained through eigen-computations.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Recovering multiplication maps
A Macaulay type matrix
with right degree in L :
E
..
 .
 L

..
.
F
..
.
And one may assume that there is
no residual root on X0 = 0:



f1 , f2 

..
.



Rν
E
F
A
B




C
D
X0 E
−−−
Rν \ X0 E
Proposition
The matrix of the multiplication by L in R/(F : G )|X0 =1 is given by:
ML = (A − BD−1 C)|X0 =1 .
⇒ The roots ξ are obtained through eigen-computations.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Cylinders passing through 5 points (1)
Suppose given five points p1 , . . . , p5 in the space.
We are looking for the cylinders passing through them
(reconstruction problem in CAD).
A cylinder ⇔ a
unitary direction
→
−
p1
t = (l, m, n) ⇔
a point (l:m:n)
p2
in P2 .
p
Passing through 4
points: a
homogeneous
polynomial of
degree 3.
−
→
t
3
p4
p5
→
−
orthogonal plane to t
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Cylinders passing through 5 points (2)
We are looking for the roots of the polynomial system
C1234 (l, m, n) = C1235 (l, m, n) = 0,
Using the classical approach: 9 points, including unwanted
directions p1 p2 , p1 p3 , p2 p3 .
Let G defining these 3 points (on a line and a cubic)
The residual resultant gives the 6 points; and it is a
geometric method:
→
−
t
−
→
t
p5
p1 , p2
p4
p3
remove p1 p2
Laurent Busé
p1 , p2
p3
p5
p4
keep p1 p2
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Cylinders passing through 5 points (2)
We are looking for the roots of the polynomial system
C1234 (l, m, n) = C1235 (l, m, n) = 0,
Using the classical approach: 9 points, including unwanted
directions p1 p2 , p1 p3 , p2 p3 .
Let G defining these 3 points (on a line and a cubic)
The residual resultant gives the 6 points; and it is a
geometric method:
→
−
t
−
→
t
p5
p1 , p2
p4
p3
remove p1 p2
Laurent Busé
p1 , p2
p3
p5
p4
keep p1 p2
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Cylinders passing through 5 points (2)
We are looking for the roots of the polynomial system
C1234 (l, m, n) = C1235 (l, m, n) = 0,
Using the classical approach: 9 points, including unwanted
directions p1 p2 , p1 p3 , p2 p3 .
Let G defining these 3 points (on a line and a cubic)
The residual resultant gives the 6 points; and it is a
geometric method:
→
−
t
−
→
t
p5
p1 , p2
p4
p3
remove p1 p2
Laurent Busé
p1 , p2
p3
p5
p4
keep p1 p2
Elimination Theory, Commutative Algebra and Applications
Classical approach
Residual approach
The residual resultant
Solving polynomial systems
An example: cylinders passing through 5 points
Cylinders passing through 5 points (2)
We are looking for the roots of the polynomial system
C1234 (l, m, n) = C1235 (l, m, n) = 0,
Using the classical approach: 9 points, including unwanted
directions p1 p2 , p1 p3 , p2 p3 .
Let G defining these 3 points (on a line and a cubic)
The residual resultant gives the 6 points; and it is a
geometric method:
→
−
t
−
→
t
p5
p1 , p2
p4
p3
remove p1 p2
Laurent Busé
p1 , p2
p3
p5
p4
keep p1 p2
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Part II
Implicitization problem
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
The implicitization problem for curves
Suppose given a generically finite rational map
φ
P1
−
→
P2
x := (x0 : x1 ) 7→ (f1 (x) : f2 (x) : f3 (x)),
where f1 (x), f2 (x), f3 (x) are homogeneous of the same degree d.
Its closed image is an irreducible curve C in P2 . We would like to
compute its implicit equation that we will denote C (T1 , T2 , T3 ).
Proposition (degree of C)
d − deg(gcd(f1 , f2 , f3 )) =
0
deg(φ)deg(C)
Laurent Busé
if φ not gen. finite,
otherwise.
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
The implicitization problem for curves
Suppose given a generically finite rational map
φ
P1
−
→
P2
x := (x0 : x1 ) 7→ (f1 (x) : f2 (x) : f3 (x)),
where f1 (x), f2 (x), f3 (x) are homogeneous of the same degree d.
Its closed image is an irreducible curve C in P2 . We would like to
compute its implicit equation that we will denote C (T1 , T2 , T3 ).
Proposition (degree of C)
d − deg(gcd(f1 , f2 , f3 )) =
0
deg(φ)deg(C)
Laurent Busé
if φ not gen. finite,
otherwise.
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
Implicitizing with resultants
Assume that there is no base point, which means here that the
gcd(f1 , f2 , f3 ) is a constant. Then:
Proposition
Res(f1 (x) − T1 f3 (x), f2 (x) − T2 f3 (x)) = C (T1 , T2 , 1)deg(φ) .
Here Res denotes the Sylvester resultant in x = (x0 , x1 ).
It can be computed as the determinant of a square matrix.
Observe that we get a universal formula
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95)
A moving line of degree ν following C is
H(x; T1 , T2 , T3 ) := g1 (x)T1 + g2 (x)T2 + g3 (x)T3
P
such that degx (H) = ν and 3i=1 gi (x)fi (x) = 0.
Assume that there is no base point,
then there exists d linearly independant moving lines of degree
d − 1, say H (1) , . . . , H (d) .
P
(i)
rewrite H (i) = νj=0 Lj (T1 , T2 , T3 )x0j x1ν−j ,
(i)
Construct the square matrix M := (Lj )
det(M) = C (T1 , T2 , T3 )deg(φ)
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95)
A moving line of degree ν following C is
H(x; T1 , T2 , T3 ) := g1 (x)T1 + g2 (x)T2 + g3 (x)T3
P
such that degx (H) = ν and 3i=1 gi (x)fi (x) = 0.
Assume that there is no base point,
then there exists d linearly independant moving lines of degree
d − 1, say H (1) , . . . , H (d) .
P
(i)
rewrite H (i) = νj=0 Lj (T1 , T2 , T3 )x0j x1ν−j ,
(i)
Construct the square matrix M := (Lj )
det(M) = C (T1 , T2 , T3 )deg(φ)
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95)
A moving line of degree ν following C is
H(x; T1 , T2 , T3 ) := g1 (x)T1 + g2 (x)T2 + g3 (x)T3
P
such that degx (H) = ν and 3i=1 gi (x)fi (x) = 0.
Assume that there is no base point,
then there exists d linearly independant moving lines of degree
d − 1, say H (1) , . . . , H (d) .
P
(i)
rewrite H (i) = νj=0 Lj (T1 , T2 , T3 )x0j x1ν−j ,
(i)
Construct the square matrix M := (Lj )
det(M) = C (T1 , T2 , T3 )deg(φ)
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95)
A moving line of degree ν following C is
H(x; T1 , T2 , T3 ) := g1 (x)T1 + g2 (x)T2 + g3 (x)T3
P
such that degx (H) = ν and 3i=1 gi (x)fi (x) = 0.
Assume that there is no base point,
then there exists d linearly independant moving lines of degree
d − 1, say H (1) , . . . , H (d) .
P
(i)
rewrite H (i) = νj=0 Lj (T1 , T2 , T3 )x0j x1ν−j ,
(i)
Construct the square matrix M := (Lj )
det(M) = C (T1 , T2 , T3 )deg(φ)
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
Implicitizing with “moving lines”
Definition (Sederberg-Chen ’95)
A moving line of degree ν following C is
H(x; T1 , T2 , T3 ) := g1 (x)T1 + g2 (x)T2 + g3 (x)T3
P
such that degx (H) = ν and 3i=1 gi (x)fi (x) = 0.
Assume that there is no base point,
then there exists d linearly independant moving lines of degree
d − 1, say H (1) , . . . , H (d) .
P
(i)
rewrite H (i) = νj=0 Lj (T1 , T2 , T3 )x0j x1ν−j ,
(i)
Construct the square matrix M := (Lj )
det(M) = C (T1 , T2 , T3 )deg(φ)
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Using resultants
Using moving lines
Using syzygies
Implicitization of curves
Implicitization of surfaces
Commutative algebra revisiting moving lines
Set A := K[x0 , x1 ][T1 , T2 , T3 ] bigraded: A(n; m).
Consider both Koszul complexes
d
d
d1 =(f1 f2 f3 )
3
2
0 → A(−3d; 0) −→
A(−2d; 0)3 −→
A(−d; 0)3 −−−−−−−−→ A(0; 0),
v
v1 =(T1 T2 T3 )
v
3
2
0 → A(0; −3) −→
A(0; −2)3 −→
A(0; −1)3 −−−−−−−−−→ A(0; 0),
they induce the following bigraded complex of A-modules:
v
v
v
3
2
1
Ker(d3 ) −→
Ker(d2 )(0; −2) −→
Ker(d1 )(0; −1) −→
A(0; 0)
P3
=0
(g1 , g2 , g3 )
7→
i=1 gi (x)Ti .
called the Z-approximation complex (Simis-Vasconcelos ’81).
Moving lines following C are exactly v1 (Ker(d1 )).
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Using resultants
Using moving lines
Using syzygies
Implicitization of curves
Implicitization of surfaces
Commutative algebra revisiting moving lines
Set A := K[x0 , x1 ][T1 , T2 , T3 ] bigraded: A(n; m).
Consider both Koszul complexes
d
d
d1 =(f1 f2 f3 )
3
2
0 → A(−3d; 0) −→
A(−2d; 0)3 −→
A(−d; 0)3 −−−−−−−−→ A(0; 0),
v
v1 =(T1 T2 T3 )
v
3
2
0 → A(0; −3) −→
A(0; −2)3 −→
A(0; −1)3 −−−−−−−−−→ A(0; 0),
they induce the following bigraded complex of A-modules:
v
v
v
3
2
1
Ker(d3 ) −→
Ker(d2 )(0; −2) −→
Ker(d1 )(0; −1) −→
A(0; 0)
P3
=0
(g1 , g2 , g3 )
7→
i=1 gi (x)Ti .
called the Z-approximation complex (Simis-Vasconcelos ’81).
Moving lines following C are exactly v1 (Ker(d1 )).
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Using resultants
Using moving lines
Using syzygies
Implicitization of curves
Implicitization of surfaces
Commutative algebra revisiting moving lines
Set A := K[x0 , x1 ][T1 , T2 , T3 ] bigraded: A(n; m).
Consider both Koszul complexes
d
d
d1 =(f1 f2 f3 )
3
2
0 → A(−3d; 0) −→
A(−2d; 0)3 −→
A(−d; 0)3 −−−−−−−−→ A(0; 0),
v
v1 =(T1 T2 T3 )
v
3
2
0 → A(0; −3) −→
A(0; −2)3 −→
A(0; −1)3 −−−−−−−−−→ A(0; 0),
they induce the following bigraded complex of A-modules:
v
v
v
3
2
1
Ker(d3 ) −→
Ker(d2 )(0; −2) −→
Ker(d1 )(0; −1) −→
A(0; 0)
P3
=0
(g1 , g2 , g3 )
7→
i=1 gi (x)Ti .
called the Z-approximation complex (Simis-Vasconcelos ’81).
Moving lines following C are exactly v1 (Ker(d1 )).
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
Implicitizing with syzygies
Theorem
The determinant of the complex of free K[T1 , T2 , T3 ]-modules
v
v
2
1
0 → Ker(d2 )[ν] (−2) −→
Ker(d1 )[ν] (−1) −→
A[ν] (0),
equals C (T1 , T2 , T3 )deg(φ) for all ν ≥ d − 1.
In general C deg(φ) is represented either as
1 (T1 ,T2 ,T3 )
• a quotient D
D2 (T1 ,T2 ,T3 ) .
• the gcd of the maximal minors of the first map.
If there is no base point then Ker(d2 )(0; 0) ' A(−d; 0), and
v1
C (T1 , T2 , T3 )deg(φ) = det Ker(d1 )[d−1] (−1) −→
A[d−1] (0) ,
which is nothing but the moving lines matrix.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
Implicitizing with syzygies
Theorem
The determinant of the complex of free K[T1 , T2 , T3 ]-modules
v
v
2
1
0 → Ker(d2 )[ν] (−2) −→
Ker(d1 )[ν] (−1) −→
A[ν] (0),
equals C (T1 , T2 , T3 )deg(φ) for all ν ≥ d − 1.
In general C deg(φ) is represented either as
1 (T1 ,T2 ,T3 )
• a quotient D
D2 (T1 ,T2 ,T3 ) .
• the gcd of the maximal minors of the first map.
If there is no base point then Ker(d2 )(0; 0) ' A(−d; 0), and
v1
C (T1 , T2 , T3 )deg(φ) = det Ker(d1 )[d−1] (−1) −→
A[d−1] (0) ,
which is nothing but the moving lines matrix.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using moving lines
Using syzygies
Implicitizing with syzygies
Theorem
The determinant of the complex of free K[T1 , T2 , T3 ]-modules
v
v
2
1
0 → Ker(d2 )[ν] (−2) −→
Ker(d1 )[ν] (−1) −→
A[ν] (0),
equals C (T1 , T2 , T3 )deg(φ) for all ν ≥ d − 1.
In general C deg(φ) is represented either as
1 (T1 ,T2 ,T3 )
• a quotient D
D2 (T1 ,T2 ,T3 ) .
• the gcd of the maximal minors of the first map.
If there is no base point then Ker(d2 )(0; 0) ' A(−d; 0), and
v1
C (T1 , T2 , T3 )deg(φ) = det Ker(d1 )[d−1] (−1) −→
A[d−1] (0) ,
which is nothing but the moving lines matrix.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
The implicitization problem for surfaces
Suppose given a generically finite rational map
φ
P2
−
→
P3
x := (x0 : x1 : x2 ) 7→ (f1 (x) : f2 (x) : f3 (x) : f4 (x))
where f1 (x), . . . , f4 (x) are homogeneous of the same degree d.
Its closed image is an irreducible surface S: S(T1 , T2 , T3 , T4 ).
Proposition (degree of S)
Assume that V (f1 , f2 , f3 , f4 ) is finite (base points are isolated):
X
0
if φ not generically finite,
2
d −
ep =
deg(φ)deg(S) otherwise.
p∈V (f1 ,f2 ,f3 ,f4 )
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
The implicitization problem for surfaces
Suppose given a generically finite rational map
φ
P2
−
→
P3
x := (x0 : x1 : x2 ) 7→ (f1 (x) : f2 (x) : f3 (x) : f4 (x))
where f1 (x), . . . , f4 (x) are homogeneous of the same degree d.
Its closed image is an irreducible surface S: S(T1 , T2 , T3 , T4 ).
Proposition (degree of S)
Assume that V (f1 , f2 , f3 , f4 ) is finite (base points are isolated):
X
0
if φ not generically finite,
2
d −
ep =
deg(φ)deg(S) otherwise.
p∈V (f1 ,f2 ,f3 ,f4 )
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
Implicitization formulas with resultants
If there is no base point then
Res(f1 (x)−T1 f4 (x), f2 (x)−T2 f4 (x), f3 (x)−T3 f4 (x)) = S(T1 , T2 , T3 , 1)deg(φ)
If there are base points then we can do similarly with
A sparse resultant (no base points in a certain toric variety)
A residual resultant (l.c.i. base points in P2 )
. . . your own ad hoc resultant.
Remark: These formulas are universal for certain classes of
parameterizations.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
Implicitization formulas with resultants
If there is no base point then
Res(f1 (x)−T1 f4 (x), f2 (x)−T2 f4 (x), f3 (x)−T3 f4 (x)) = S(T1 , T2 , T3 , 1)deg(φ)
If there are base points then we can do similarly with
A sparse resultant (no base points in a certain toric variety)
A residual resultant (l.c.i. base points in P2 )
. . . your own ad hoc resultant.
Remark: These formulas are universal for certain classes of
parameterizations.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
Implicitization formulas with resultants
If there is no base point then
Res(f1 (x)−T1 f4 (x), f2 (x)−T2 f4 (x), f3 (x)−T3 f4 (x)) = S(T1 , T2 , T3 , 1)deg(φ)
If there are base points then we can do similarly with
A sparse resultant (no base points in a certain toric variety)
A residual resultant (l.c.i. base points in P2 )
. . . your own ad hoc resultant.
Remark: These formulas are universal for certain classes of
parameterizations.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Using resultants
Using syzygies
Implicitization of curves
Implicitization of surfaces
Implicitizing with syzygies (1)
Set A := K[x0 , x1 , x2 ][T1 , T2 , T3 , T4 ] bigraded: A(n; m).
As for curve implicitization, consider both Koszul complexes
d
d
(f1 ,...,f4 )
d
4
3
2
0 → A(−4d; 0) −→
A(−3d; 0)4 −→
A(−2d; 0)6 −→
A(−d; 0)4 −−−−−→ A(0; 0),
v
v
v1 =(T1 ,...,T4 )
v
4
3
2
0 → A(0; −4) −→
A(0; −3)4 −→
A(0; −2)6 −→
A(0; −1)4 −−−−−−−−→ A(0; 0).
They induce the following bigraded complex of A-modules:
v
v
v
4
3
2
Ker(d4 ) = 0 −→
Ker(d3 )(0; −3) −→
Ker(d2 )(0; −2) −→
v
1
Ker(d1 )(0; −1) −→
A(0; 0)
P4
(g1 , g2 , g3 , g4 ) 7→
i=1 gi (x)Ti (x).
Moving planes following S are exactly v1 (Ker(d1 ))
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
Implicitizing with syzygies (2)
Set I := (f1 , f2 , f3 , f4 ) ⊂ K[x0 , x1 , x2 ].
Theorem
Assume V (I ) ⊂ P2 is finite and an almost l.c.i. (locally at most 3
generators). Then, for all integers
ν ≥ ν0 := 2(d − 1) − indeg(I sat ),
the determinant of the complex of free K[T1 , . . . , T4 ]-modules
v
v
v
3
2
1
0 → Ker(d3 )[ν] (−3) −→
Ker(d2 )[ν] (−2) −→
Ker(d1 )[ν] (−1) −→
A[ν] (0),
is homogeneous of degree d 2 −
P
is a multiple of S(T1 , T2 , T3 , T4
p∈V (I ) dp ,
deg(φ)
)
,
is exactly S deg(φ) if and only if I is a l.c.i..
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
Implicitizing with syzygies (3)
⇒ In general S deg(φ) is represented either as
• a quotient
D1 (T1 ,T2 ,T3 ,T4 ).D3 (T1 ,T2 ,T3 ,T4 )
.
D2 (T1 ,T2 ,T3 ,T4 )
• the gcd of the maximal minors of the first map v1 .
⇒ The bound ν0 = 2(d − 1) − indeg(I sat ):
indeg(I sat ) := min{t ∈ N : Itsat 6= 0}. It is a geometric invariant of
the scheme V (I ): the smallest degree of a (non-zero) hypersurface
passing through all the points defined by I . We have:
• ν0 = 2d − 2 if I has no base points (I sat = K[x0 , x1 , x2 ]),
• 2d − 3 ≥ ν0 ≥ d − 2 if I has base points,
• ν0 = d − 2 if I is saturated (and we also have D3 = 1).
Remark: The presence of base points simplifies the problem!
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
Implicitizing with syzygies (3)
⇒ In general S deg(φ) is represented either as
• a quotient
D1 (T1 ,T2 ,T3 ,T4 ).D3 (T1 ,T2 ,T3 ,T4 )
.
D2 (T1 ,T2 ,T3 ,T4 )
• the gcd of the maximal minors of the first map v1 .
⇒ The bound ν0 = 2(d − 1) − indeg(I sat ):
indeg(I sat ) := min{t ∈ N : Itsat 6= 0}. It is a geometric invariant of
the scheme V (I ): the smallest degree of a (non-zero) hypersurface
passing through all the points defined by I . We have:
• ν0 = 2d − 2 if I has no base points (I sat = K[x0 , x1 , x2 ]),
• 2d − 3 ≥ ν0 ≥ d − 2 if I has base points,
• ν0 = d − 2 if I is saturated (and we also have D3 = 1).
Remark: The presence of base points simplifies the problem!
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
Implicitizing with syzygies (3)
⇒ In general S deg(φ) is represented either as
• a quotient
D1 (T1 ,T2 ,T3 ,T4 ).D3 (T1 ,T2 ,T3 ,T4 )
.
D2 (T1 ,T2 ,T3 ,T4 )
• the gcd of the maximal minors of the first map v1 .
⇒ The bound ν0 = 2(d − 1) − indeg(I sat ):
indeg(I sat ) := min{t ∈ N : Itsat 6= 0}. It is a geometric invariant of
the scheme V (I ): the smallest degree of a (non-zero) hypersurface
passing through all the points defined by I . We have:
• ν0 = 2d − 2 if I has no base points (I sat = K[x0 , x1 , x2 ]),
• 2d − 3 ≥ ν0 ≥ d − 2 if I has base points,
• ν0 = d − 2 if I is saturated (and we also have D3 = 1).
Remark: The presence of base points simplifies the problem!
Laurent Busé
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
Examples
• A cubic with 6 base points (Sederberg-Chen):

f1



f2
f3



f4
=
=
=
=
s 2 t + 2t 3 + s 2 u + 4stu + 4t 2 u + 3su 2 + 2tu 2 + 2u 3 ,
−s 3 − 2st 2 − 2s 2 u − stu + su 2 − 2tu 2 + 2u 3 ,
−s 3 − 2s 2 t − 3st 2 − 3s 2 u − 3stu + 2t 2 u − 2su 2 − 2tu 2 ,
s 3 + s 2 t + t 3 + s 2 u + t 2 u − su 2 − tu 2 − u 3 .
I is saturated ⇒ ν0 = d − 2 = 1. The surface is represented by
the matrix:


x
 y
z
−z − w
x − 2y + z − 2w
−x − 2w
Laurent Busé
y +w
2y − z  .
y + 2w
Elimination Theory, Commutative Algebra and Applications
Implicitization of curves
Implicitization of surfaces
Using resultants
Using syzygies
Examples
• An example with a fat base point:

f1



f2
f3



f4
=
=
=
=
s 3 − 6s 2 t − 5st 2 − 4s 2 u + 4stu − 3t 2 u,
−s 3 − 2s 2 t − st 2 − 5s 2 u − 3stu − 6t 2 u,
−4s 3 − 2s 2 t + 4st 2 − 6t 3 + 6s 2 u − 6stu − 2t 2 u,
2s 3 − 6s 2 t + 3st 2 − 6t 3 − 3s 2 u − 4stu + 2t 2 u.
I defines exactly the fat point p = (s, t)2 (dp = 3 and ep = 4).
The method gives a multiple of degree d 2 − dp = 9 − 3 = 6 (and
not 5).
Here ν0 = 2(3 − 1) − 2 = 2, we obtain a single 6 × 6 matrix.
Laurent Busé
Elimination Theory, Commutative Algebra and Applications