Elimination Theory, Commutative Algebra and Applications · Outlines Elimination Theory, Commutative Algebra and Applications Laurent Bus´e INRIA Sophia-Antipolis, France March 2,

Outlines

Elimination Theory, Commutative Algebra andApplications

Laurent Buse

INRIA Sophia-Antipolis, France

March 2, 2005

Laurent Buse Elimination Theory, Commutative Algebra and Applications

OutlinesPart I: Residual resultants in P2

Part II: Implicitization Problem

Part I: Solving polynomial systems with resultants

1 Overview of the classical approachThe Macaulay resultantSolving zero-dimensional polynomial systemLimitations

2 The residual resultant approachThe residual resultantSolving polynomial systemsAn example: cylinders passing through 5 points


OutlinesPart I: Residual resultants in P2

Part II: Implicitization Problem

Part II: Implicitization problem for curves and surfaces

3 Implicitization of rational plane curvesUsing resultantsUsing moving linesUsing syzygies

4 Implicitization of rational surfacesUsing resultantsUsing syzygies


Classical approachResidual approach

Part I

Solving polynomial systems with resultants



The Macaulay resultantSolving zero-dimensional polynomial systemLimitations

The Macaulay resultant

Consider a system of 3 homogeneous polynomials in X0,X1,X2:f0 =

∑|α|=d0

c0,α.Xα

f1 =∑|α|=d1

c1,α.Xα

f2 =∑|α|=d2

c2,α.Xα

Xα denotes a monomial Xα00 Xα1

1 Xα22 (αi ≥ 0).

The ci,α’s are the parameters of the system with value in analgebraically closed field k.

Theorem (Macaulay - 1902)

There exists an irreducible and homogeneous polynomial ink[ci,α], denoted Res(f0, f1, f2), satisfying:

Res(f0, f1, f2) = 0 ⇔ ∃x ∈ P2k such that f0(x) = f1(x) = f2(x) = 0.




The Macaulay resultant

Consider a system of 3 homogeneous polynomials in X0,X1,X2:f0 =

∑|α|=d0

c0,α.Xα

f1 =∑|α|=d1

c1,α.Xα

f2 =∑|α|=d2

c2,α.Xα

Xα denotes a monomial Xα00 Xα1

1 Xα22 (αi ≥ 0).

The ci,α’s are the parameters of the system with value in analgebraically closed field k.

Theorem (Macaulay - 1902)

There exists an irreducible and homogeneous polynomial ink[ci,α], denoted Res(f0, f1, f2), satisfying:

Res(f0, f1, f2) = 0 ⇔ ∃x ∈ P2k such that f0(x) = f1(x) = f2(x) = 0.




Computing the Macaulay resultant

Degrees: Res(f0, f1, f2) is homogeneous w.r.t. each fi :

degci,α=

d0d1d2

diwith i ∈ {0, 1, 2} fixed.

Computation: Let ν := d0 + d1 + d2 − 2, set A := k[ci,α]and consider the map (the first map of a Koszul complex)

3⊕i=1

A[X0,X1,X2]ν−di → A[X0,X1,X2]ν

(g0, g1, g2) 7→ g0f0 + g1f1 + g2f2.

It depends on the ci,α’s and has maximal rankSome maximal minors are the Macaulay matrices.Their gcd (3 are sufficients) gives Res(f0, f1, f2).






degci,α=

d0d1d2



3⊕i=1


(g0, g1, g2) 7→ g0f0 + g1f1 + g2f2.







degci,α=

d0d1d2



3⊕i=1


(g0, g1, g2) 7→ g0f0 + g1f1 + g2f2.





Recovering the Chow form

Let f1, f2 ∈ R = k[X0,X1,X2] be homo-geneous polynomials defining isolatedpoints (a complete intersection) in P2.

Introduce L(X) = c0X0 + c1X1 + c2X2.L(X )

V (f1, f2)

Proposition (Chow form)

Res(L, f1, f2) =∏

ξ∈V (f1,f2)

L(ξ)µ(ξ),

where µ(ξ) denotes the multiplicity (or degree) of ξ.

⇒ An Absolute factorisation problem





Let f1, f2 ∈ R = k[X0,X1,X2] be homo-geneous polynomials defining isolatedpoints (a complete intersection) in P2.

Introduce L(X) = c0X0 + c1X1 + c2X2.L(X )

V (f1, f2)


Res(L, f1, f2) =∏

ξ∈V (f1,f2)

L(ξ)µ(ξ),


⇒ An Absolute factorisation problem




Recovering multiplication maps

The Macaulay matrix withright degree in L :

E F...

...L f1, f2...

...

Rν

And one may assume that there isno solutions on X0 = 0:

E FA B

C D

X0E−−−Rν \ X0E

Proposition

The matrix of the multiplication by L in R/(f1, f2)|X0=1 is given by:

ML = (A− BD−1C)|X0=1.

⇒ The roots ξ are obtained through eigen-computations.






E F...

...L f1, f2...

...

Rν


E FA B

C D


Proposition


ML = (A− BD−1C)|X0=1.







E F...

...L f1, f2...

...

Rν


E FA B

C D


Proposition


ML = (A− BD−1C)|X0=1.





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 ofthe parameters ci ,α’s) then Res(L, f1, f2) is also identicallyzero.

Questions:

How to remove the non finite part of V (f1, f2)?

More generally, how to compute “a part” (evenzero-dimensional) of V (f1, f2)?




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 ofthe parameters ci ,α’s) then Res(L, f1, f2) is also identicallyzero.

Questions:

How to remove the non finite part of V (f1, f2)?

More generally, how to compute “a part” (evenzero-dimensional) of V (f1, f2)?




Three circles in the projective plane

Consider the following system:f0 = c0,1X

20 + c0,2X0X1 + c0,3X0X2 + c0,4(X

21 + X 2

2 )f1 = c1,1X

20 + c1,2X0X1 + c1,3X0X2 + c1,4(X

21 + X 2

2 )f2 = c2,1X

20 + c2,2X0X1 + c2,3X0X2 + c2,4(X

21 + X 2

2 )

Problem: When do they have a common point ?

f1

f2

P2

f0

Two “base points”: (0 : 1 : ±i).This implies Res(f0, f1, f2) ≡ 0.

Refined question: When do theyhave a common root which is not(0 : 1 : ±i) ?






20 + c0,2X0X1 + c0,3X0X2 + c0,4(X

21 + X 2

2 )f1 = c1,1X

20 + c1,2X0X1 + c1,3X0X2 + c1,4(X

21 + X 2

2 )f2 = c2,1X

20 + c2,2X0X1 + c2,3X0X2 + c2,4(X

21 + X 2

2 )


f1

f2

P2

f0








20 + c0,2X0X1 + c0,3X0X2 + c0,4(X

21 + X 2

2 )f1 = c1,1X

20 + c1,2X0X1 + c1,3X0X2 + c1,4(X

21 + X 2

2 )f2 = c2,1X

20 + c2,2X0X1 + c2,3X0X2 + c2,4(X

21 + X 2

2 )


f1

f2

P2

f0





The residual resultantSolving polynomial systemsAn 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 ofpositive degree k1, . . . , km, respectively.Let d0,d1,d2 be integers such that di ≥ kj for all (i , j).

One considers the system :f0 =

∑mi=1 hi,0(X) gi(X) total degree d0

f1 =∑m

i=1 hi,1(X) gi(X) total degree d1

f2 =∑m


where hi ,j(X ) =∑|α|=dj−ki

ci,jα Xα.

Main fact

The Polynomials fi are generic of degree di in the idealG = (g1, . . . , gm).

Rk : The case G = (1) corresponds to the previous setting.Laurent Buse Elimination Theory, Commutative Algebra and Applications



The setting





f1 =∑m


f2 =∑m



ci,jα Xα.

Main fact





The setting





f1 =∑m


f2 =∑m



ci,jα Xα.

Main fact





The setting





f1 =∑m


f2 =∑m



ci,jα Xα.

Main fact





Existence

Notations :

F := (f0, f1, f2) ⊂ A[X0,X1,X2] with A = k[ci,jα ],

(F : G ) = {f ∈ A[X0,X1,X2] such that fG ⊂ F}.

Theorem

If G is a local complete intersection and max(di ) ≥ min(kj) + 1then 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.




Existence

Notations :



Theorem


α → k

Res(f0, f1, f2) = 0 ⇔ V (F : G ) 6= ∅.


V (F : R) = V (F ) 6= ∅ ⇔ ∃x ∈ Pn : f0(x) = f1(x) = f2(x) = 0.




Existence

Notations :



Theorem


α → k

Res(f0, f1, f2) = 0 ⇔ V (F : G ) 6= ∅.


V (F : R) = V (F ) 6= ∅ ⇔ ∃x ∈ Pn : f0(x) = f1(x) = f2(x) = 0.




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⊕i=1

R(−li )φ−→

m⊕i=1

R(−ki )(g1,...,gm)−−−−−−→ G → 0.

L := Mat(φ) =

p1,1 . . . p1,m−1...

...pm,1 . . . pm,m−1

.

Proposition (multi-degree)

The degree is completely explicit in the di ’s, kj ’s and lt ’s:

degfi (Res(f0, f1, f2)) =d0d1d2

di−

∑n−1i=1 l2j −

∑ni=1 k2

j

2.





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⊕i=1

R(−li )φ−→

m⊕i=1

R(−ki )(g1,...,gm)−−−−−−→ G → 0.

L := Mat(φ) =

p1,1 . . . p1,m−1...

...pm,1 . . . pm,m−1

.

Proposition (multi-degree)

The degree is completely explicit in the di ’s, kj ’s and lt ’s:

degfi (Res(f0, f1, f2)) =d0d1d2

di−

∑n−1i=1 l2j −

∑ni=1 k2

j

2.





Construct the m × (m + 2) matrix gluing L and the hi ,j ’s:

L|H :=

p1,1 . . . p1,m−1 h1,0 h1,1 h1,2

......

......

...pm,1 . . . pm,m−1 hm,0 hm,1 hm,2

,

Set ν := d0 + d1 + d2 − 2− 2min kj and let I ⊂ Nm be theset indexing the m ×m minors of L|H. Construct the matrixM of the map (first map of an Eagon-Northcott complex):⊕

s∈I

A[X0,X1,X2]ν−deg(∆s) → A[X0,X1,X2]ν

(. . . , bs , . . .) 7→∑s∈I

bs∆s

Proposition

This matrix “represents” Res(f0, f1, f2).






L|H :=

p1,1 . . . p1,m−1 h1,0 h1,1 h1,2

......

......

...pm,1 . . . pm,m−1 hm,0 hm,1 hm,2

,


s∈I


(. . . , bs , . . .) 7→∑s∈I

bs∆s

Proposition







L|H :=

p1,1 . . . p1,m−1 h1,0 h1,1 h1,2

......

......

...pm,1 . . . pm,m−1 hm,0 hm,1 hm,2

,


s∈I


(. . . , bs , . . .) 7→∑s∈I

bs∆s

Proposition






Let f1, f2 ∈ R = k[X0,X1,X2] be homo-geneous polynomials defining some iso-lated points Z and V (G) (l.c.i. codim2 ACM) in P2.

Introduce L(X) = c0X0 + c1X1 + c2X2

and choose f vanishing on V (G ) and noton Z. L(X )

V (f1, f2)

V (G)


Res(Lf , f1, f2) =∏

ξ∈V (F :G)

L(ξ)µ(ξ),


⇒ An absolute factorisation problem





Let f1, f2 ∈ R = k[X0,X1,X2] be homo-geneous polynomials defining some iso-lated points Z and V (G) (l.c.i. codim2 ACM) in P2.

Introduce L(X) = c0X0 + c1X1 + c2X2

and choose f vanishing on V (G ) and noton Z. L(X )

V (f1, f2)

V (G)


Res(Lf , f1, f2) =∏

ξ∈V (F :G)

L(ξ)µ(ξ),


⇒ An absolute factorisation problem





A Macaulay type matrixwith right degree in L :

E F...

...L f1, f2...

...

Rν

And one may assume that there isno residual root on X0 = 0:

E FA B

C D


Proposition

The matrix of the multiplication by L in R/(F : G )|X0=1 is given by:

ML = (A− BD−1C)|X0=1.







E F...

...L f1, f2...

...

Rν


E FA B

C D


Proposition


ML = (A− BD−1C)|X0=1.







E F...

...L f1, f2...

...

Rν


E FA B

C D


Proposition


ML = (A− BD−1C)|X0=1.





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 ⇔ aunitary direction−→t = (l ,m, n) ⇔a point (l:m:n)in P2.

Passing through 4points: ahomogeneouspolynomial ofdegree 3.

−→t

orthogonal plane to−→t

p1

p2

p3

p4

p5





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 unwanteddirections p1p2, p1p3, p2p3.

Let G defining these 3 points (on a line and a cubic)The residual resultant gives the 6 points; and it is ageometric method:

p1, p2

p3p4

p5

p3

p1, p2

p4

p5

−→t

−→t

remove p1p2 keep p1p2






C1234(l ,m, n) = C1235(l ,m, n) = 0,



p1, p2

p3p4

p5

p3

p1, p2

p4

p5

−→t

−→t







C1234(l ,m, n) = C1235(l ,m, n) = 0,



p1, p2

p3p4

p5

p3

p1, p2

p4

p5

−→t

−→t







C1234(l ,m, n) = C1235(l ,m, n) = 0,



p1, p2

p3p4

p5

p3

p1, p2

p4

p5

−→t

−→t



Implicitization of curvesImplicitization of surfaces

Part II

Implicitization problem



Using resultantsUsing moving linesUsing syzygies

The implicitization problem for curves

Suppose given a generically finite rational map

P1 φ−→ 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 tocompute its implicit equation that we will denote C (T1,T2,T3).

Proposition (degree of C)

d − deg(gcd(f1, f2, f3)) =

{0 if φ not gen. finite,

deg(φ)deg(C) otherwise.




The implicitization problem for curves


P1 φ−→ 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 tocompute its implicit equation that we will denote C (T1,T2,T3).

Proposition (degree of C)

d − deg(gcd(f1, f2, f3)) =

{0 if φ not gen. finite,

deg(φ)deg(C) otherwise.




Implicitizing with resultants

Assume that there is no base point, which means here that thegcd(f1, f2, f3) is a constant. Then:

Proposition

Res(f1(x)− T1f3(x), f2(x)− T2f3(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




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

such that degx(H) = ν and∑3

i=1 gi (x)fi (x) = 0.

Assume that there is no base point,

then there exists d linearly independant moving lines of degreed− 1, say H(1), . . . ,H(d).

rewrite H(i) =∑ν

j=0 L(i)j (T1,T2,T3)x

j0x

ν−j1 ,

Construct the square matrix M := (L(i)j )

det(M) = C (T1,T2,T3)deg(φ)







H(x;T1,T2,T3) := g1(x)T1 + g2(x)T2 + g3(x)T3


i=1 gi (x)fi (x) = 0.



rewrite H(i) =∑ν

j=0 L(i)j (T1,T2,T3)x

j0x

ν−j1 ,









H(x;T1,T2,T3) := g1(x)T1 + g2(x)T2 + g3(x)T3


i=1 gi (x)fi (x) = 0.



rewrite H(i) =∑ν

j=0 L(i)j (T1,T2,T3)x

j0x

ν−j1 ,









H(x;T1,T2,T3) := g1(x)T1 + g2(x)T2 + g3(x)T3


i=1 gi (x)fi (x) = 0.



rewrite H(i) =∑ν

j=0 L(i)j (T1,T2,T3)x

j0x

ν−j1 ,









H(x;T1,T2,T3) := g1(x)T1 + g2(x)T2 + g3(x)T3


i=1 gi (x)fi (x) = 0.



rewrite H(i) =∑ν

j=0 L(i)j (T1,T2,T3)x

j0x

ν−j1 ,






Commutative algebra revisiting moving lines

Set A := K[x0, x1][T1,T2,T3] bigraded: A(n;m).

Consider both Koszul complexes

0 → A(−3d ; 0)d3−→ A(−2d ; 0)3

d2−→ A(−d ; 0)3d1=(f1 f2 f3)−−−−−−−−→ A(0; 0),

0 → A(0;−3)v3−→ A(0;−2)3

v2−→ A(0;−1)3v1=(T1 T2 T3)−−−−−−−−−→ A(0; 0),

they induce the following bigraded complex of A-modules:

Ker(d3)v3−→ Ker(d2)(0;−2)

v2−→ Ker(d1)(0;−1)v1−→ A(0; 0)

= 0 (g1, g2, g3) 7→∑3

i=1 gi (x)Ti .

called the Z-approximation complex (Simis-Vasconcelos ’81).

Moving lines following C are exactly v1(Ker(d1)).







0 → A(−3d ; 0)d3−→ A(−2d ; 0)3

d2−→ A(−d ; 0)3d1=(f1 f2 f3)−−−−−−−−→ A(0; 0),

0 → A(0;−3)v3−→ A(0;−2)3

v2−→ A(0;−1)3v1=(T1 T2 T3)−−−−−−−−−→ A(0; 0),


Ker(d3)v3−→ Ker(d2)(0;−2)

v2−→ Ker(d1)(0;−1)v1−→ A(0; 0)

= 0 (g1, g2, g3) 7→∑3

i=1 gi (x)Ti .









0 → A(−3d ; 0)d3−→ A(−2d ; 0)3

d2−→ A(−d ; 0)3d1=(f1 f2 f3)−−−−−−−−→ A(0; 0),

0 → A(0;−3)v3−→ A(0;−2)3

v2−→ A(0;−1)3v1=(T1 T2 T3)−−−−−−−−−→ A(0; 0),


Ker(d3)v3−→ Ker(d2)(0;−2)

v2−→ Ker(d1)(0;−1)v1−→ A(0; 0)

= 0 (g1, g2, g3) 7→∑3

i=1 gi (x)Ti .






Implicitizing with syzygies

Theorem

The determinant of the complex of free K[T1,T2,T3]-modules

0 → Ker(d2)[ν](−2)v2−→ Ker(d1)[ν](−1)

v1−→ A[ν](0),

equals C (T1,T2,T3)deg(φ) for all ν ≥ d − 1.

In general Cdeg(φ) is represented either as

• a quotient D1(T1,T2,T3)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

C (T1,T2,T3)deg(φ) = det

(Ker(d1)[d−1](−1)

v1−→ A[d−1](0))

,

which is nothing but the moving lines matrix.





Theorem


0 → Ker(d2)[ν](−2)v2−→ Ker(d1)[ν](−1)

v1−→ A[ν](0),




.




(Ker(d1)[d−1](−1)

v1−→ A[d−1](0))

,






Theorem


0 → Ker(d2)[ν](−2)v2−→ Ker(d1)[ν](−1)

v1−→ A[ν](0),




.




(Ker(d1)[d−1](−1)

v1−→ A[d−1](0))

,




Using resultantsUsing syzygies

The implicitization problem for surfaces


P2 φ−→ 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):

d2−∑

p∈V (f1,f2,f3,f4)

ep =

{0 if φ not generically finite,

deg(φ)deg(S) otherwise.




The implicitization problem for surfaces


P2 φ−→ 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):

d2−∑

p∈V (f1,f2,f3,f4)

ep =

{0 if φ not generically finite,

deg(φ)deg(S) otherwise.




Implicitization formulas with resultants

If there is no base point then

Res(f1(x)−T1f4(x), f2(x)−T2f4(x), f3(x)−T3f4(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 ofparameterizations.






















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

0 → A(−4d ; 0)d4−→ A(−3d ; 0)4

d3−→ A(−2d ; 0)6d2−→ A(−d ; 0)4

(f1,...,f4)−−−−−→ A(0; 0),

0 → A(0;−4)v4−→ A(0;−3)4

v3−→ A(0;−2)6v2−→ A(0;−1)4

v1=(T1,...,T4)−−−−−−−−→ A(0; 0).

They induce the following bigraded complex of A-modules:

Ker(d4) = 0v4−→ Ker(d3)(0;−3)

v3−→ Ker(d2)(0;−2)v2−→

Ker(d1)(0;−1)v1−→ A(0; 0)

(g1, g2, g3, g4) 7→∑4

i=1 gi (x)Ti (x).

Moving planes following S are exactly v1(Ker(d1))





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 3generators). Then, for all integers

ν ≥ ν0 := 2(d − 1)− indeg(I sat),

the determinant of the complex of free K[T1, . . . ,T4]-modules

0 → Ker(d3)[ν](−3)v3−→ Ker(d2)[ν](−2)

v2−→ Ker(d1)[ν](−1)v1−→ A[ν](0),

is homogeneous of degree d2 −∑

p∈V (I ) dp,

is a multiple of S(T1,T2,T3,T4)deg(φ),

is exactly Sdeg(φ) if and only if I is a l.c.i..





⇒ In general Sdeg(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 : I sat

t 6= 0}. It is a geometric invariant ofthe scheme V (I ): the smallest degree of a (non-zero) hypersurfacepassing 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!







.














.











Examples

• A cubic with 6 base points (Sederberg-Chen):f1 = s2t + 2t3 + s2u + 4stu + 4t2u + 3su2 + 2tu2 + 2u3,f2 = −s3 − 2st2 − 2s2u − stu + su2 − 2tu2 + 2u3,f3 = −s3 − 2s2t − 3st2 − 3s2u − 3stu + 2t2u − 2su2 − 2tu2,f4 = s3 + s2t + t3 + s2u + t2u − su2 − tu2 − u3.

I is saturated ⇒ ν0 = d − 2 = 1. The surface is represented bythe matrix: x −z − w y + w

y x − 2y + z − 2w 2y − zz −x − 2w y + 2w

.




Examples

• An example with a fat base point:f1 = s3 − 6s2t − 5st2 − 4s2u + 4stu − 3t2u,f2 = −s3 − 2s2t − st2 − 5s2u − 3stu − 6t2u,f3 = −4s3 − 2s2t + 4st2 − 6t3 + 6s2u − 6stu − 2t2u,f4 = 2s3 − 6s2t + 3st2 − 6t3 − 3s2u − 4stu + 2t2u.

I defines exactly the fat point p = (s, t)2 (dp = 3 and ep = 4).The method gives a multiple of degree d2 − dp = 9− 3 = 6 (andnot 5).Here ν0 = 2(3− 1)− 2 = 2, we obtain a single 6× 6 matrix.


Documents

Elimination Theory, Commutative Algebra and Applications · Outlines Elimination Theory, Commutative Algebra and Applications Laurent Bus´e INRIA Sophia-Antipolis, France March 2,