1_841_wang.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 3, No. 4, 2010, 602-632

ISSN 1307-5543 – www.ejpam.com

Minimal Generators for the Rees Algebra of Rational Space

Curves of Type (1, 1, d − 2)

J. William Hoffman1, Haohao Wang2,∗, Xiaohong Jia3, Ron Goldman4

1 Department of Mathematics, Louisiana State University, Baton Rouge, LA, USA
2 Department of Mathematics, Southeast Missouri State University, Cape Girardeau, MO, USA
3 Department of Computer Science, The University of Hong Kong, Hong Kong, P. R. China
4 Computer Science Department, Rice University, 6100 Main St., MS-132, Houston, TX 77251,

USA

Abstract. We provide an algorithm to find a minimal set of generators for the Rees algebra associated

to rational space curves of type (1,1, d−2) in projective 3-space based solely on a µ-basis of the curve.

We also illustrate the geometry behind the generators via a case study of rational quartic space curves.

2000 Mathematics Subject Classifications: 14Q05 (primary), 13D02, 14H20(secondary)

Key Words and Phrases: Rees Algebra, Syzygy, Implicit Equations, Resultant, µ-basis

1. Introduction

The Rees algebra of an ideal I ⊂ R defined as the graded algebra (with the elements of R

having degree 0 and the elements of I having degree 1)

Rees(I) = R⊕ I ⊕ I2 ⊕ ...

is a classical algebraic structure which has been studied for decades by the Commutative

Algebra community, see [25]. One motivation for this study is that it is related to a classical

problem in elimination theory: the implicitization problem. The implicitization problem is to

find an algorithm to convert a parametrization given by a rational map

f : Pm→ Pn

into defining equations for the closure X of the image f (Pm). Rational curves and surfaces

are widely used in Computer Aided Design, since it is easy to describe the points on these

∗Corresponding author.

Email addresses: hoffman�math.lsu.edu (J. Hoffman), hwang�semo.edu (H. Wang), xhjia�
s.hku.hk
(X. Jia), rng�ri
e.edu (R. Goldman)

http://www.ejpam.com 602 c© 2010 EJPAM All rights reserved.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 603

curves and surfaces by means of their parameter values. However, it is not convenient to use

the parametric representations to describe the set of points that are common to two different

parametrically defined curves or surfaces. Thus there is a need to go back and forth between

a parametric and an implicit description of a curve or surface. This is, in essence, the implic-

itization problem: to develop efficient algorithms to generate implicit equations for a curve

or surface for which one knows a parametric representation. Implicitization algorithms have

been most highly developed in the case that X is a hypersurface, so X is defined by only one

equation F = 0. It turns out that F is related to the structure of the Rees algebra of I .

For instance consider the implicitization problem for rational surfaces in P3. Algebraically

the problem is this: Given an ideal I = 〈 f0, f1, f2, f3〉 ⊂ R where the fi are homogeneous

polynomials of degree d in the standard Z-graded ring R := K[s, t,u] over an infinite field

K, find a minimal set of generators for the kernel of the map h: R[x0, x1, x2, x3]→ Rees(I),

where h(x i) = fi for i = 0,1,2,3. Under certain general circumstances, the implicit equation

F is the element of ker(h) of degree 0 in the variables s, t,u.

Elements of ker(h), under the name of moving lines and moving planes, were introduced

into Computer Aided Geometric Design by Sederberg, Cox and their collaborators in order

to develop robust, efficient algorithms for implicitizing rational curves and surfaces [10],

[20], [21], [22]. In the past two years, [4], [7], [8], and [16] utilized the method of moving

curves and surfaces to determine the defining equations for the Rees algebra of plane algebraic

curves. They each develop different methods and algorithms for finding explicit moving curves

that are a minimal set of generators for the associated Rees algebra. The approach in [7], [8]

is based on iterations of Sylvester determinants, regular sequences, and local cohomology

computations

This approach to the implicitization problem, which has been developed especially by

Jouanoulou, Busé, Chardin, Cox, D’Andrea and others, utilizes the structure of a free resolu-

tion of I as an R-module (see [2], [3]). However, these studies have been largely limited to

the case of hypersurface parametrizations, and they lead to expressions for F as determinants

of certain complexes. See also [1], [4], [5], [15].

The corresponding problems for codimension two (and higher) are much more difficult.

Given an ideal I = 〈 f0, f1, . . . , fn〉 ⊂ R of height two where f0, f1, . . . , fn are homogeneous

polynomials of degree d in the standard Z-graded ring R := K[s, t] over an infinite field K,

the goal is to find a minimal set of generators for the kernel of the map h: R[x0, x1, . . . , xn]→
Rees(I), where h(x i) = fi for i = 0,1, . . . , n.

The first nontrivial case is that of rational space curves, n = 3. Finding a minimal set of

generators for the Rees algebra of the ideal I of a rational space curve solves the implicitization

problem for space curves. Since the ideal I gives rise to a rational function P1→ Pn mapping

(s, t)→ (x0, . . . , xn), the image of this map is a curve C ⊂ Pn with homogeneous coordinate

ring K[x0, . . . , xn]. The ideal theoretic implicit equations of the curve C are among the

minimal generators of the defining equations for the Rees algebra associated to the space

curve. For example, [6, 11, 12, 19, 23], have all investigated minimal generators for the Rees

algebra of the ideal of a rational space curve.

Recently, Kustin, Polini and Ulrich [18] studied minimal generators of the defining equa-

tions for the Rees algebra of a height two ideal I with a minimal free resolution of the follow-



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 604

ing form:

0→ R(−d − 1)n−1
⊕

R(−2d + n+ 1) −−−→ Rn(−d)
[ f0,..., fn]−−−−→ I → 0.

Since the homogeneous coordinate ring K[x0, . . . , xn] is isomorphic to the special fiber ring of

I , their approach to determining the generators for the Rees algebra of the curve C is to find

the defining ideal of this special fiber ring. They pay close attention to the depth and algebraic

properties of this fiber ring of I , and give an explicit description of the minimal generators.

In this paper, we specialize the setting in Kustin, Polini and Ulrich [18] to rational space

curves of type (1,1, d−2) in projective 3-space. We do not prove any new theorems about the

structure of Rees algebras; rather our primary goal is to provide an algorithmic approach and

to give elementary constructions for the minimal generators for the Rees algebra associated to

these curves based solely on their µ-basis. Our approach is to study separately the cases when

the rational curve is either singular or non-singular. If the rational space curve is singular,

then we study separately the cases when the degree of the curve is either even or odd. The

generators of the Rees algebra are all expressed entirely in terms of the three elements of the

µ-basis. We will prove our results by comparing the generators produced by our algorithm

with those described in [18]. Our algorithm shows that the very complicated description of

the generators of the Rees algebra given in [18] can be simplified considerably in the case

of rational space curves of type (1,1, d − 2). The second goal of this paper is to illustrate

the geometry behind the generators via a case study of rational quartic space curves. We will

construct the implicit equations of the curve and the defining equations for the Rees algebra

in a simple manner from the elements of the µ-basis.

We proceed in the following fashion. In Section 2 we review the basic notion of moving

surfaces and µ-bases, and recall some results concerning how to detect the singularities of

rational space curves using µ-bases. In Section 3 we provide an algorithm to find minimal

generators for the Rees algebra of the ideal of a rational space curve based solely on the three

elements of a µ-basis of the rational space curve. In Section 4 we illustrate the geometry

behind the generators with a case study of rational quartic space curves.

2. Moving Planes and µ-bases

Throughout this paper, we shall consider rational space curves C in three-dimensional

projective space over a field K of characteristic 0, given as the image of a generic 1-1 rational

parametrization:

F(s, t) = ( f0(s, t), f1(s, t), f2(s, t), f3(s, t)), (s, t) 6= (0,0), (1)

where f0, f1, f2, f3 are linearly independent homogeneous polynomials of the same degree

d ≥ 3 in the standard Z-graded ring R := K[s, t], and gcd( f0, f1, f2, f3) = 1.

We begin by briefly recalling some basic definitions.

Definition 1. A moving surface of degree r is a polynomial
∑

i+ j+ℓ+k=r

Ai jℓk(s, t)x i y jzℓwk, Ai jℓk ∈ R.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 605

This polynomial is said to follow the parametrization (1) if

∑

i+ j+ℓ+k=r

Ai jℓk(s, t) f0(s, t)i f1(s, t) j f2(s, t)ℓ f3(s, t)k ≡ 0.

Hence a moving surface of degree r follows the parametrization (1) if and only if

(Ai jℓk)i+ j+ℓ+k=r ∈ Syz(I r ),

where I is the ideal 〈 f0, f1, f2, f3〉 ⊂ R. The set of all moving surfaces that follow the parametriza-

tion (1) is an ideal in R[x , y, z, w], called the moving surface ideal.

Remark 1. If f is a moving surface, then f is in the bigraded ringK[s, t; x , y, z, w] = R[x , y, z, w],

and deg( f ) = (d1, d2) where d1 is the degree in s, t, and d2 is the degree in x , y, z, w.

Definition 2. A moving plane is a moving surface of degree d2 = 1. An axial moving plane

is a moving plane where all the planes of the family pass through either a common point A or a

common line
←→
AB. The point A is called an axis point, and the line

←→
AB is called the axis line or

axis of the moving plane. Similarly, a moving quadric is a moving surface of degree d2 = 2.

When we refer to the degree of a moving plane (or moving quadric), we are referring to the degree

in s, t, i.e., d1.

For example, consider the rational quintic space curve where

( f0, f1, f2, f3) = (s
5, s3 t2, s2 t3, t5). The polynomial t3 x−s3z is an axial moving plane of degree

3 that follows the curve with axis
←→
AB where A= (0,0,0,1) and B = (0,1,0,1).

Remark 2. A moving plane of degree one in s, t always has an axis line; a moving plane of degree

two in s, t always has an axis point, and may have an axis line.

Indeed if we write a moving plane of degree one in s, t as

f (x , y, z, w)s + g(x , y, z, w)t, where deg( f ) = deg(g) = 1, gcd( f , g) = 1,

then the variety V( f , g) is the axis of the moving plane. It is easy to see that V( f , g) is a linear

variety of dimension at least one; hence a moving plane of degree one in s, t always has an axis

line.

In the language of Commutative Algebra, the collection of moving planes that follow

a parametrization F(s, t) = ( f0, f1, f2, f3) is exactly Syz(I), where I = 〈 f0, f1, f2, f3〉. The

Hilbert-Burch theorem [9, Chapter 6] says that the minimal free resolution of the ideal I has

the following form:

0→ R(−d −µ1)⊕ R(−d −µ2)⊕ R(−d −µ3)
p,q,r
−−−→ R4(−d)

f0, f1, f2, f3
−−−−−→ I → 0,

where µ1 ≤ µ2 ≤ µ3, µ1 + µ2 + µ3 = d = deg( fi). Thus the syzygies Syz( f0, f1, f2, f3) are

generated by p,q, r. The generators p,q, r are called a µ-basis. We sometimes write p,q, r as



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 606

three independent moving planes p,q, r, where p = p·X, q = q·X, r = r·X and X= (x , y, z, w),

of homogeneous degrees µ1 ≤ µ2 ≤ µ3 in s, t:

p = px x + py y + pzz + pww,

q = qx x + qy y + qzz + qww,

r = rx x + ry y + rzz + rww,

and these three elements p,q, r are also called a µ-basis. There is a simple algorithm to

compute a µ-basis from the polynomials f0, f1, f2, f3 using only Gaussian elimination [24].

Although the µ-basis elements are not unique, the degrees µ1,µ2,µ3 of the µ-basis elements

are unique. From now on, we will use (µ1,µ2,µ3) to denote the type of a rational space curve.

Throughout this paper, we focus on rational space curves of type (1,1, d − 2).

Here we recall two results concerning the relation between µ-bases and points on rational

space curves.

Proposition 1. [[26]] Let F(s, t) be a rational space curve with a µ-basis p,q, r. Then a point

Q is on the space curve F(s, t) if and only if

deg(gcd(p ·Q, q ·Q, r ·Q))≥ 1.

Moreover, the roots of this gcd are the parameters with proper multiplicity corresponding to the

point Q on the curve F(s, t).

Proposition 2. [[26]] Suppose p(s, t), q(s, t) and r(s, t) are a µ-basis of degrees (1,1, d − 2)

for the rational space curve F(s, t). Then

1. F(s, t) has no singularities if and only if the axes of p and q do not intersect.

2. F(s, t) has exactly one singular point A which is of order d − 2 if and only if the axes of p

and q intersect at the point A.

Moreover, from two µ-basis elements, we can generate a quadric surface that contains a

space curve of type (1,1, d − 2).

Lemma 1. Let p = p1s+ p0 t and q = q1s+q0t be two µ-basis elements of degree 1 of a rational

quartic space curve C . Then the curve C is contained in the irreducible quadric surface defined

by Sylvs,t(p,q) = det

�

p1 p0

q1 q0

�

= 0.

Proof. Since deg(p) = deg(q) = (1,1), p and q are relatively prime over the ring C . Hence

p and q form a regular sequence over C , and therefore Sylvs,t(p,q) 6≡ 0. Moreover, since

deg(pi) = deg(qi) = 1 for i = 0,1, Sylvs,t(p,q) is of homogeneous degree 2. In addition,

Sylvs,t(p,q) vanishes on the curve C , since p,q are two moving planes that follow the curve.

Finally, this quadric is irreducible, otherwise the curve C would be contained in one of the

linear factors, contradicting the fact that the curve C is non-planar.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 607

3. Minimal Generators of the Rees Algebra

In this section, we will investigate minimal generators for the Rees algebra associated to

rational space curves of type (1,1, d − 2). We will provide a simple algorithm to find these

generators based solely on the three µ-basis elements of these rational space curves. Our

approach is to study separately the cases when the curve F(s, t) is either singular or non-

singular. If F(s, t) is singular, then we also study separately the cases when the degree of the

curve is either even or odd.

3.1. Singular Rational Space Curves of Type (1, 1, d − 2)

Here we will find a minimal set of generators for the Rees algebra associated to singular

rational space curves of type (1,1, d − 2).

Remark 3. Let p = p1s+ p0 t and q = q1s+ q0t be two µ-basis elements of a singular rational

space curve F(s, t) of type (1,1, d − 2), where p1, p0,q1,q0 are linear forms in x , y, z, w. Then

the polynomials p1, p0,q1,q0 are linearly dependent of rank 3.

Proof. By Proposition 2, there is a unique singular point on the curve, which is the inter-

section of the axes of p and q. Thus, V(p1, p0,q1,q0) consists of just one point. Therefore, the

polynomials p1, p0,q1,q0 are linearly dependent of rank 3.

Lemma 2. Let p,q, r be a µ-basis for a singular rational space curve F(s, t) of type (1,1, d − 2).

By a linear transformation on the basis elements p,q, and by a projective change of coordinates

in x , y, z, w, we can adjust the elements of the µ-basis so that p = ys − x t, q = zs − y t, and

transform the singular point to (0,0,0,1).

Proof. Let p = p1s+ p0 t and q = q1s+q0 t where p1, p0,q1,q0 are linear forms in x , y, z, w.

Since p,q are µ-basis elements, gcd(p1, p0) = gcd(q1,q0) = 1; otherwise the curve would be

contained in the plane p1 = p0 = 0 or q1 = q0 = 0. Also, note that gcd(p1,q1) = gcd(p0,q0) =

1. Otherwise, if q1 = ap1 for some non-zero constant a, then q = q1s + q0 t = ap1s + q0t,

and q− ap = (ap1s+ q0t)− a(p1s+ p0 t) = (q0 − ap0)t. But this is impossible, since q− ap

is a moving plane that follows the space curve F(s, t), so the space curve F(s, t) would be

contained in the plane q0 − ap0. Thus gcd(p1,q1) = 1. Similarly, gcd(p0,q0) = 1.

Since gcd(q1,q0) = 1, by a change of coordinates we can let y ′ = q1 and x ′ = −q0. Hence

q = y ′s − x ′t. Since gcd(p1,q1) = 1, it follows that p1 6= a y ′ for any non-zero constant. We

will now prove the lemma by establishing the following two cases.

Case 1: p1 = ax ′+ b y ′ for some a, b ∈K and a 6= 0. Then p = (ax ′+ b y ′)s+p0t. Therefore, the

moving plane 1

a
p− b

a
q = x ′s+

p0+bx ′

a
t is linearly independent from the moving plane q,

and follows the space curve F(s, t). Thus, q, 1

a
p − b

a
q and r also form a µ-basis for the

space curve F(s, t). Since the space curve F(s, t) is singular, the axes of q and 1

a
p− b

a
q

intersect at one point. Thus, by Remark 3, x ′, y ′,
p0+bx ′

a
must have rank 3. Hence, by

the projective change of coordinates, x = − p0+bx ′

a
, y = x ′, z = y ′, we obtain a µ-basis

with elements p = ys− x t and q = zs− y t.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 608

Case 2: p1 6= ax ′ + b y ′ for all possible a, b ∈ K. Then p1 is linearly independent of x ′, y ′.

Without loss of generality, we can let z′ = p1. Then p = z′s+ p0t. Since the space curve

F(s, t) is singular, by Remark 3 x ′, y ′, z′, p0 has rank 3, so p0 = ax ′+ b y ′+ cz′ for some

non-zero constant a, b, c. Moreover, gcd(p0,q0) = 1 implies that b, c cannot both be

zero. Now consider the two moving planes

bq+ cp = b(y ′s− x ′ t) + c[z′s+ (ax ′+ b y ′ + cz′)t]

= (b y ′ + cz′)s+ [(ac − b)x ′+ bc y ′ + c2z′]t,

−aq− p = −a(y ′s− x ′ t)− [z′s+ (ax ′+ b y ′ + cz′)t]

= (−a y ′ − z′)s− (b y ′ + cz′)t.

Notice that ac − b 6= 0, otherwise the moving plane bq + cp = (b y ′ + cz′)s + [bc y ′ +

c2z′]t = (b y ′+ cz′)(s+ ct), which is not possible because the curve F(s, t) is not planar.

Thus bq+ cp and −aq− p are two linearly independent moving planes that follow the

space curve F(s, t). Therefore, by a projective change of coordinates, if we let x =

−[(ac − b)x ′ + bc y ′ + c2z′], y = b y ′ + cz′, z = −a y ′ − z′, then we obtain a µ-basis in

the desired form: p = ys− x t and q = zs− y t.

Finally, by Proposition 2, the singular point is (0,0,0,1).

By Lemma 2, we can choose the µ-basis elements so that tp + sq = zs2 − x t2. Therefore,

since the µ-basis vanishes on the curve, we have the following relations for any point on the

space curve F(s, t)

t

s
=

y

x
,

t2

s2
=

z

x
,

s

t
=

y

z
,

s2

t2
=

x

z
. (2)

We will find the implicit equations of these space curves, and also the defining equations

for the Rees algebras of these space curve by studying separately the case when d is even and

the case when d is odd.

3.1.1. Degree d = 2k, k ≥ 2

First, observe that when d = 2k, the µ-basis element r can be written as

r = rd−2sd−2 + rd−3sd−3 t + · · ·+ r0 td−2 =

d−2
∑

j=0

r js
j td−2− j = s2i[

d−2
∑

j=0

r j

s j td−2− j

s2i
]

= s2i[

2i−1
∑

j=0

r j t
d−2−2i(

t

s
)2i− j +

d−2
∑

j=2i

r js
j−2i td−2− j]

= s2i{
i−1
∑

j=0

[r2 j t
d−2−2i(

t2

s2
)i− j + r2 j+1 td−2−2i(

t2

s2
)i− j−1(

t

s
)] +

d−2−2i
∑

j=0

r2i+ js
j td−2−2i− j},

For all i = 1, . . . , k− 1, let

r ′i (A, B) =

i−1
∑

j=0

[r2 j t
d−2−2i(B)i− j + r2 j+1 td−2−2i(B)i− j−1(A)]+

d−2−2i
∑

j=0

r2i+ js
j td−2−2i− j .



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 609

Then

r ′i (
y

x
,

z

x
) = td−2−2i

i−1
∑

j=0

[r2 j(
z

x
)i− j + r2 j+1(

z

x
)i− j−1(

y

x
)] +

d−2−2i
∑

j=0

r2i+ js
j td−2−2i− j . (3)

Theorem 1. A minimal set of generators for the defining equation of the Rees algebra associated

to a singular rational space curve of type (1,1, d − 2) where d = 2k are given by the following

k+ 3 polynomials:

1. three µ-basis elements: p, q, r, where deg(p) = deg(q) = (1,1), and deg(r) = (d − 2,1);

2. two implicit equations: Sylvs,t(p,q) of degree (0,2), and x k−1r ′
k−1
(

y

x
, z

x
) of degree (0, k);

3. k− 2 moving surfaces: x i r ′
i
(

y

x
, z

x
) of degree (d − 2− 2i, i + 1) for i = 1, . . . , k− 2;

where r ′i (
y

x
, z

x
) is defined as in Equation (3) for i = 1, . . . k− 1.

Proof. We will apply the results of Kustin, Polini and Ulrich [18], by listing the minimal set

of generators in their paper, and comparing these generators with the generators listed in our

theorem.

First, we note that the notation x , y, z, w, t, s, ri , i = 0, . . . , d − 2 in this paper is the same

as T1, T2, T3, T4, x , y, ci , i = 0, . . . , d−2 in their notation. Moreover, in our setting, we identify

the following items in their paper for singular curves of even degrees:

ρ = 1, ℓ = 2, σ1 = 2, σ2 = 1, T2,1 = s, T2,2 = t, T1,1 = x , T1,2 = y, T1,3 = z.

By Theorem 3.2 in [18], we have

a= (a1) = 0,1, . . . , k− 2; f (a) = f (a1) = d − 3− 2a1;

r(a) = r(a1) = 1; f (;) = k− 2; r(;) = 1; T ; = 1.

By definition 3.5 and the description in [18], the generators for the Rees algebra are

p,q, Sylvs,t(p,q), f1 and ga1,1. We shall now write f1 and ga1,1 explicitly and compare these

expressions with the generators listed in the statement of our theorem. We have

f1 = z
∑

i+ j=k−3

x iz j(r2iz + r2i+1 y) + x k−2(rd−2 x + rd−3 y + rd−4z)

=
∑

i+ j=k−3

[r2i x
iz j+2 + r2i+1 x i yz j+1] + (rd−2 x k−1+ rd−3 x k−2 y + rd−4 x k−2z)

= x k−1{rd−2 + rd−3(
y

x
) + rd−4(

z

x
) +

k−3
∑

i=0

[r2i(
z

x
)k−1−i + r2i+1(

z

x
)k−2−i(

y

x
)]}

= x k−1{rd−2 +

k−2
∑

i=0

[r2i(
z

x
)k−1−i + r2i+1(

z

x
)k−2−i(

y

x
)]}



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 610

= x k−1r ′k−1(
y

x
,

z

x
).

Hence f1 is one of our implicit equations and deg( f1) = (0, k).

ga1,1 = td−2−2a1

∑

i+ j=a1−1

x iz j(r2iz + r2i+1 y) + x a1 t ×

∑

i+ j=d−4−2a1

si t j r2a1+i t + x a1sd−3−2a1(rd−2s+ rd−3 t)

= x a1{td−2−2a1

a1−1
∑

i=0

[r2i(
z

x
)a1−i + r2i+1(

z

x
)a1−i−1(

y

x
)] +

d−2−2a1
∑

i=0

r2a1+is
i td−2−2a1−i}

=

(

x a1 r ′a1
(

y

x
, z

x
) for a1 = 1, · · · , k− 2,

r for a1 = 0.

Hence ga1,1 are our moving surfaces, and deg(ga1,1) = (d − 2− 2a1, a1 + 1). Therefore, the

generators provided by our theorem are the same as the generators described in [18]. Thus,

we have proved our claim.

3.1.2. Degree d = 2k+ 1, k ≥ 2

First, observe that when d = 2k+ 1, the µ-basis element r can be written as

r = rd−2sd−2 + rd−3sd−3 t + · · ·+ r0 td−2 =

d−2
∑

j=0

r js
j td−2− j = sd−2

d−2
∑

j=0

r j

td−2− j

sd−2− j

= sd−2
k−1
∑

j=0

[r2 j(
t2

s2
)k−1− j(

t

s
) + r2 j+1(

t2

s2
)k−1− j].

Let

r ′′(A, B) =

k−1
∑

j=0

[r2 j(B)
k−1− j(A) + r2 j+1(B)

k−1− j].

Then

r ′′(
y

x
,

z

x
) =

k−1
∑

j=0

[r2 j(
z

x
)k−1− j(

y

x
) + r2 j+1(

z

x
)k−1− j]. (4)

On the other hand, the µ-basis element r also can be written as

r = rd−2sd−2 + rd−3sd−3t + · · ·+ r0 td−2 =

d−2
∑

j=0

r js
j td−2− j = td−2

d−2
∑

j=0

r j

s j

t j

= td−2{
k−1
∑

j=0

[r2 j(
s2

t2
) j + r2 j+1(

s2

t2
) j(

s

t
)]}.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 611

Let

r ′′′(A, B) =

k−1
∑

j=0

[r2 j(B)
j + r2 j+1(B)

j(A)];

then

r ′′′(
y

z
,

x

z
) =

k−1
∑

j=0

[r2 j(
x

z
) j + r2 j+1(

x

z
) j(

y

z
)]. (5)

Moreover, for all i = 1, . . . , k− 1,

r = rd−2sd−2 + rd−3sd−3 t + · · ·+ r0 td−2 =

d−2
∑

j=0

r js
j td−2− j = s2i[

d−2
∑

j=0

r j

s j td−2− j

s2i
]

= s2i[

2i−1
∑

j=0

r j t
d−2−2i(

t

s
)2i− j +

d−2
∑

j=2i

r js
j−2i td−2− j]

= s2i{
i−1
∑

j=0

[r2 j t
d−2−2i(

t2

s2
)i− j + r2 j+1 td−2−2i(

t2

s2
)i− j−1(

t

s
)] +

d−2−2i
∑

j=0

r2i+ js
j td−2−2i− j}.

For all i = 1, . . . , k− 1, let

r ′′i (A, B) = td−2−2i
i−1
∑

j=0

[r2 j(B)
i− j + r2 j+1(B)

i− j−1(A)]+

d−2−2i
∑

j=0

r2i+ js
j td−2−2i− j ;

then

r ′′i (
y

x
,

z

x
) = td−2−2i

i−1
∑

j=0

[r2 j(
z

x
)i− j + r2 j+1(

z

x
)i− j−1(

y

x
)] +

d−2−2i
∑

j=0

r2i+ js
j td−2−2i− j . (6)

Theorem 2. A minimal set of generators for the Rees algebra associated to a singular rational

space curve of type (1,1, d − 2) where d = 2k+ 1 are given by the following k+ 5 polynomials:

1. three µ-basis elements: p, q, r, where deg(p) = deg(q) = (1,1) and deg(r) = (d − 2,1);

2. three implicit equations: Sylvs,t(p,q) of degree (0,2), x kr ′′(
y

x
, z

x
) and zkr ′′′(

y

z
, x

z
) of de-

gree (0, k+ 1);

3. k− 1 moving surfaces: x i r ′′
i
(

y

x
, z

x
) of degree (d − 2− 2i, i + 1) for i = 1, . . . , k− 1;

where r ′′(
y

x
, z

x
), r ′′′(

y

z
, x

z
) and r ′′i (

y

x
, z

x
) are defined in Equations (4), (5) and (6).

Proof. We will apply the results of Kustin, Polini and Ulrich [18], by listing the set of

minimal generators in their paper, and comparing these generators with the generators listed

in the statement of our theorem.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 612

First, we note that the notation x , y, z, w, t, s, ri , i = 0, . . . , d − 2 in this paper is the same

as T1, T2, T3, T4, x , y, ci , i = 0, . . . , d−2 in their notation. Moreover, in our setting, we identify

the following items in their paper for singular curves of odd degrees:

ρ = 1, ℓ = 2, σ1 = 2, σ2 = 1, s = T2,1, t = T2,2, T1,1 = x , T1,2 = y, T1,3 = z.

By Theorem 3.2 in [18],

a= (a1) = 0,1, . . . , k− 1; f (a) = f (a1) = d − 3− 2a1;

r(a) = r(a1) = 1; f (;) = k− 1; r(;) = 2; T ; = 1.

By definition 3.5 and the description in [18], the generators for the Rees algebra are

p,q, Sylvs,t(p,q), f1 and ga1,1. We shall now write f1 and ga1,1 explicitly and compare these

expressions with the generators listed in the statement of our theorem.

f1 = y
∑

i+ j=k−2

x iz j(r2iz + r2i+1 y) + x k−1(rd−2 x + rd−3 y)

=

k−2
∑

i=0

[r2i x
izk−1 y + r2i+1 x izk−2−i y2] + rd−2 x k + rd−3 x k y

= x k{
k−2
∑

i=0

[r2i

zk−1 y

x k−i
+ r2i+1

zk−2−i y2

x k−i
] + rd−2 + rd−3

y

x
}

= x k
k−1
∑

i=0

[r2i(
z

x
)k−1−i(

y

x
) + r2i+1(

z

x
)k−1−i], since y2 = xz on the curve

= x kr ′′(
y

x
,

z

x
).

Hence f1 is one of our implicit equations and deg( f1) = (0, k+ 1).

f2 = z
∑

i+ j=k−2

x iz j(r2iz + r2i+1 y) + x k−1(rd−2 y + rd−3z)

=

k−2
∑

i=0

[r2i x
izk−i + r2i+1 x izk−1−i y] + rd−2 x k−1 y + rd−3 x k−1z

= zk{
k−2
∑

i=0

[r2i(
x

z
)i + r2i+1(

x

z
)i(

y

z
)] + rd−2(

x

z
)k−1(

y

z
) + rd−3(

x

z
)k−1}

= zk
k−1
∑

i=0

[r2i(
x

z
)i + r2i+1(

x

z
)i(

y

z
)]

= zkr ′′′(
y

z
,

x

z
).

Hence f2 is another one of our implicit equations and deg( f2) = (0, k+ 1).

ga1,1 = td−2−2a1

∑

i+ j=a1−1

x iz j(r2iz + r2i+1 y) + x a1 t ×



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 613

∑

i+ j=d−4−2a1

si t j r2a1+i t + x a1sd−3−2a1(rd−2s+ rd−3 t)

= x a1{td−2−2a1

a1−1
∑

i=0

[r2i(
z

x
)a1−i + r2i+1(

z

x
)a1−1−i(

y

x
)] +

d−2−2a1
∑

i=0

r2a1+is
i td−2−2a1−i}

ga1,1 =

(

x a1 r ′′a1
(

y

x
, z

x
) for a1 = 1, · · · , k− 1,

r for a1 = 0.

Hence ga1,1 are our moving surfaces and deg(ga1,1) = (d − 2− 2a1, a1 + 1). Therefore, the

generators provided by our theorem are the same as the generators described in [18]. Thus,

we have proved our claim.

3.2. Non-Singular Rational Space Curve of Type (1, 1, d − 2)

Now, we will find a minimal set of generators for the Rees algebra associated to non-

singular rational space curves of type (1,1, d − 2).

Lemma 3. Let p,q, r be a µ-basis for a non-singular rational space curve of type (1,1, d − 2).

By a linear transformation on the basis elements p,q, and by a projective change of coordinates

in x , y, z, w, we can adjust the elements of the µ-basis so that p = ys− x t, q = ws− zt.

Proof. First write p = p1s + p0 t and q = q1s+ q0 t where p1, p0,q1,q0 are linear forms in

K[x , y, z, w]. Since the curve is non-singular, by Proposition 2, V(p1, p0,q1,q0) = ;. Hence the

polynomials p1, p0,q1,q0 are linearly independent of rank 4. Thus, without loss of generality,

we can set p = ys− x t and q = ws− zt.

Now since the µ-basis vanishes on the curve, we have the following relations for any point

on the space curve F(s, t)
t

s
=

y

x
,

t

s
=

w

z
. (7)

For i = 0,1, · · · , d − 2, the µ-basis element r can be written as

r = rd−2sd−2 + rd−3sd−3 t + · · ·+ r0 td−2 =

d−2
∑

j=0

r js
j td−2− j = sd−2

d−2
∑

j=0

r j(
t

s
)d−2− j

= sd−2[

i−1
∑

j=0

r j(
t

s
)d−2−i(

t

s
)i− j +

d−2
∑

j=i

r j(
t

s
)d−2− j], where

i−1
∑

j=0

r j(
t

s
)d−2−i(

t

s
)i− j := 0 if i = 0

= sd−2[

i−1
∑

j=0

r j(
t

s
)d−2−i(

t

s
)i− j +

d−2−i
∑

j=0

ri+ j(
t

s
)d−2−i− j].

For i = 0,1, · · · , d − 2, let

r ′i (A, B) =

i−1
∑

j=0

r j(B)
d−2−i(A)i− j+

d−2−i
∑

j=0

ri+ j(B)
d−2−i− j , where

i−1
∑

j=0

r j(B)
d−2−i(A)i− j := 0 if i = 0.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 614

Then

r ′i (
y

x
,

w

z
) =

i−1
∑

j=0

r j(
w

z
)d−2−i(

y

x
)i− j +

d−2−i
∑

j=0

ri+ j(
w

z
)d−2−i− j , (8)

where
∑i−1

j=0 r j(
w

z
)d−2−i(

y

x
)i− j := 0 if i = 0.

Moreover, for all i = 1, . . . , d − 3, and for all 0≤ j ≤ i, we can also write

r = rd−2sd−2 + rd−3sd−3 t + · · ·+ r0 td−2 =

d−2
∑

ℓ=0

rℓs
ℓ td−2−ℓ

= si[

i−1
∑

ℓ=0

rℓ t
d−2−i(

t

s
)i−ℓ+

d−2
∑

ℓ=i

rℓ t
d−2−ℓsℓ−i]

= si{td−2−i[

j−1
∑

ℓ=0

rℓ(
t

s
)i− j(

t

s
) j−ℓ+

i−1
∑

ℓ= j

rℓ(
t

s
)i−ℓ] +

d−2
∑

ℓ=i

rℓ t
d−2−ℓsℓ−i},

where

i−1
∑

ℓ= j

rℓ(
t

s
)i−ℓ := 0, if j = i

= si{td−2−i[

j−1
∑

ℓ=0

rℓ(
t

s
)i− j(

t

s
) j−ℓ+

i−1− j
∑

ℓ=0

r j+ℓ(
t

s
)i− j−ℓ] +

d−2−i
∑

ℓ=0

ri+ℓ t
d−2−i−ℓsℓ}.

For all i = 1, . . . , d − 3, and for all 0≤ j ≤ i, let

r ′j,i− j(A, B) = td−2−i[

j−1
∑

ℓ=0

rℓ(B)
i− j(A) j−ℓ+

i−1− j
∑

ℓ=0

r j+ℓ(B)
i− j−ℓ] +

d−2−i
∑

ℓ=0

ri+ℓ t
d−2−i−ℓsℓ,

where
∑i−1− j

ℓ=0
r j+ℓ(B)

i− j−ℓ := 0, if j = i. Then

r ′j,i− j(
y

x
,
w

z
)

= td−2−i[

j−1
∑

ℓ=0

rℓ(
w

z
)i− j(

y

x
) j−ℓ+

i−1− j
∑

ℓ=0

r j+ℓ(
w

z
)i− j−ℓ] +

d−2−i
∑

ℓ=0

ri+ℓ t
d−2−i−ℓsℓ,

(9)

where
∑i−1− j

ℓ=0
r j+ℓ(

w

z
)i− j−ℓ := 0, if j = i.

Theorem 3. A minimal set of generators for the Rees algebra associated to a non-singular ratio-

nal space curve type (1,1, d − 2) are given by the following 3+ d +
d(d − 3)

2
polynomials:

1. three µ-basis elements: p, q, r where deg(p) = deg(q) = (1,1), and deg(r) = (d − 2,1);

2. d implicit equations: Sylvs,t(p,q) of degree 2, and d−1 implicit equations x izd−2−i r ′i (
y

x
, w

z
)

of degree d − 1 for i = 0,1, . . . , d − 2;



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 615

3.
d(d−3)

2
moving planes: x jz i− j r ′j,i− j(

y

x
, w

z
) of degree (d−2− i, i+1) for all i = 1, , . . . , d−3

and 0≤ j ≤ i;

where r ′i (
y

x
, w

z
) and r ′j,i− j(

y

x
, w

z
) are defined in Equations (8) and (9).

Proof. We will apply the results of Kustin, Polini and Ulrich [18], by listing the set of

minimal generators in their paper, and comparing these generators with the generators listed

in our theorem.

First, we note that the notation x , y, z, w, t, s, ri , i = 0, . . . , d − 2 in this paper is the same

as T1, T2, T3, T4, x , y, ci , i = 0, . . . , d−2 in their notation. Moreover, in our setting, we identify

the following items in their paper for non-singular curves:

ρ = 2, ℓ= 3, σ1 = σ2 = σ3 = 1, [T3,1, T3,2; T1,1, T1,2, T1,3; T2,1, T2,2] = [s, t; x , y, z; z, w].

By Theorem 3.2 in [18], we have

a= (a1, a2), 0≤ a1 + a2 ≤ d − 3; f (a1) = d − 3− a1; f (a1, a2) = d − 3− a1− a2;

r(a1) = 1; r(a1, a2) = 1; f (;) = d − 3; r(;) = 1; T ; = 1.

By definition 3.5 and the description in [18], the generators for the Rees algebra are

p,q, Sylvs,t(p,q), f1, ga1,1, and h(a1, a2). We shall now write f1, ga1,1, and h(a1, a2) explicitly

and compare these expressions with the generators listed in the statement of our theorem.

We have

f1 = y
∑

i+ j=d−4

x i y j(ri y) + x d−3(rd−2 x + rd−3 y) =

d−4
∑

i=0

ri x
i yd−2−i + rd−2 x d−2+ rd−3 x d−3 y

= x d−2
d−2
∑

i=0

ri(
y

x
)d−2−i = x d−2r ′d−2(

y

x
,
w

z
).

Hence f1 is one of our implicit equations and deg( f1) = (0, d − 1).

ga1,1 = wd−2−a1

∑

i+ j=a1−1

x i y j(ri y) + x a1 w ×

∑

i+ j=d−4−a1

z iw j ra1+iw + x a1zd−3−a1 (rd−2z + rd−3w)

=

a1−1
∑

i=0

ri x
i ya1−iwd−2−a1 +

d−4−a1
∑

i=0

ra1+i x
a1z iwd−2−a1−i + rd−2 x a1zd−2−a1 + rd−3 x a1zd−3−a1 w

= x a1zd−2−a1[

a1−1
∑

i=0

ri(
w

z
)d−2−a1(

y

x
)a1−i +

d−2−a1
∑

i=0

ra1+i(
w

z
)d−2−a1−i]

= x a1zd−2−a1 r ′a1
(

y

x
,

w

z
).



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 616

Hence each ga1,1 is one of our implicit equations and deg(ga1,1) = (0, d−1), a1 = 0,1, . . . , d−
3.

ha1,a2
= td−2−a1−a2[wa2

∑

i+ j=a1−1

x i y j(ri y) + x a1

∑

i+ j=a2−1

z iw j ra1+iw]

+x a1za2

d−2−a1−a2
∑

i=0

ra1+a2+i t
d−2−a1−a2−isi

= td−2−a1−a2[

a1−1
∑

i=0

ri x
i ya1−iwa2 +

a2−1
∑

i=0

ra2+i x
a1z iwa2−i]

+x a1za2

d−2−a1−a2
∑

i=0

ra1+a2+i t
d−2−a1−a2−isi

= x a1za2{td−2−a1−a2[

a1−1
∑

i=0

ri(
w

z
)a2(

y

x
)a1−i +

a2−1
∑

i=0

ra2+i(
w

z
)a2−i]

+

d−2−a1−a2
∑

i=0

ra1+a2+i t
d−2−a1−a2−isi}

=

(

x a1za2 r ′a1,a2
(

y

x
, w

z
), a1 + a2 6= 0, 0< a1 + a2 ≤ d − 3,

r a1 + a2 = 0.

Hence ha1,a2
are our moving surfaces and deg(ha1,a2

) = (d − 2− a1 − a2, a1 + a2 + 1), 0 ≤
a1 + a2 ≤ d − 3.

Therefore, the generators provided by our theorem are the same as the generators de-

scribed in [18]. Thus, we have proved our claim.

3.3. Algorithm

Based on Theorems 1, 2, and 3, we now provide a simple algorithm to find a minimal set

of generators for the Rees algebra associated to rational space curves of type (1,1, d − 2) in

projective 3-space based solely on a µ-basis of the curve.

Algorithm 4. Given a rational space curve C as the image of a generic 1-1 rational parametriza-

tion F(s, t) as in Equation (1), we compute a minimal set of generators for the associated Rees

algebra.

1. Compute a µ-basis p,q, r, and write in the following form:

p = p1s+ p0 t, q = q1s+ q0 t, r = rd−2sd−2 + rd−3sd−3 t + · · ·+ r0 td−2.

2. Compute V(p1, p0,q1,q0).



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 617

3. If V(p1, p0,q1,q0) 6= ;, and if p1 = bq1− aq0 for some a, b ∈K and a 6= 0, then let

X = −
p0 − bq0

a
, Y = −q0, Z = q1, W = w

Otherwise, find non-zero constant a, b, c ∈K, such that p0 = −aq0 + bq1+ cp1, and let

X = (ac − b)q0− bcq1 − c2p1, Y = bq1+ cp1, Z = −aq1− p1, W = w,

so that

p = Y s− X t, q = Zs− Y t, r = Rd−2sd−2 + Rd−3sd−3 t + · · ·+ R0td−2,

where Ri are linear in X , Y, Z ,W.

(1) If d = 2k, then a minimal set of generators for the associated Rees algebra is

p, q, r, Sylvs,t(p,q), X iR′i

�

Y

X
,

Z

X

�

, i = 0, . . . , k− 1,

where

X iR′i

�

Y

X
,

Z

X

�

= td−2−2i
i−1
∑

j=0

�

R2 j

�

Z

X

�i− j

+ r2 j+1

�

Z

X

�i− j−1�Y

X

�
�

+

d−2−2i
∑

j=0

R2i+ js
j td−2−2i− j .

(2) If d = 2k+ 1, then a minimal set of generators for the associated Rees algebra is

p, q, r, Sylvs,t(p,q), R′′, R′′′, X iR′i

�

Y

X
,

Z

X

�

, i = 1, . . . , k− 1,

where

R′′
�

Y

X
,

Z

X

�

=

k−1
∑

j=0

�

R2 j

�

Z

X

�k−1− j�Y

X

�

+R2 j+1

�

Z

X

�k−1− j
�

R′′′
�

Y

Z
,
X

Z

�

=

k−1
∑

j=0

�

R2 j

�

X

Z

� j

+ R2 j+1

�

X

Z

� j�Y

Z

�
�

R′′i (
Y

X
,

Z

X
) = td−2−2i

i−1
∑

j=0

�

R2 j

�

Z

X

�i− j

+ R2 j+1

�

Z

X

�i− j−1�Y

X

�
�

+

d−2−2i
∑

j=0

R2i+ js
j td−2−2i− j .



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 618

4. If V(p1, p0,q1,q0) = ;, then let X = −p0, Y = p1, Z = −q0, W = q1, so that

p = Y s− X t, q =Ws− Z t, r = Rd−2sd−2 + Rd−3sd−3 t + · · ·+ R0 td−2,

where Ri are linear in X , Y, Z ,W. A minimal set of generators for the associated Rees

algebra is

p, q, r, Sylvs,t(p,q), X ℓZ d−2−ℓR′ℓ

�

Y

X
,
W

Z

�

, X j Z i− jR′j,i− j

�

Y

X
,
W

Z

�

,

0≤ ℓ ≤ d − 2, 0≤ j ≤ i = 1, . . . , d − 3,

where

R′ℓ

�

Y

X
,
W

Z

�

=

ℓ−1
∑

j=0

R j

�

W

Z

�d−2−ℓ�Y

X

�ℓ− j

+

d−2−ℓ
∑

j=0

Rℓ+ j

�

W

Z

�d−2−ℓ− j

,

R′j,i− j

�

Y

X
,
W

Z

�

= td−2−i





j−1
∑

ℓ=0

Rℓ

�

W

Z

�i− j�Y

Z

� j−ℓ

+

i−1− j
∑

ℓ=0

R j+ℓ

�

W

Z

�i− j−ℓ




+

d−2−i
∑

ℓ=0

Ri+ℓ t
d−2−i−ℓsℓ.

Below we present two examples to show how to use Algorithm 4 to find a minimal set

of generators for the associated Rees algebra both for singular and for non-singular rational

space curves.

Example 1. Consider the rational quintic space curve given by

F(s, t) = (s4t + s3 t2 − 2s2t3, s5 + 5s4t + 6s3t2 − 4s2 t3 − 8st4, s4 t − 3s2 t3 + 2st4, t5).

Compute a µ-basis for F(s, t):

p = xs− (4z + y − 8x)t, q = (x − z)s− x t, r = (s3 + s2 t − 2st2)w + t3(z − x).

The point (0,0,0,1) is a singular point of order 3.

By a projective changes of coordinates, let

X = −8x + y + 4z, Y = x , Z = x − z, W = w.

Then the new µ-basis can be written as

p = Y s− X t, q = Zs− Y t, r =Ws3 +Ws2 t − 2Wst2 − Z t3.

The minimal generators for the Rees algebra associated to this curve are p,q, r and the following

polynomials:

Sylvs,t(p,q) = X Z − Y 2 = −9x2+ x y + 12xz− yz − 4z2;



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 619

X 2r ′′(
Y

X
,

Z

X
) = X 2{[1+ (

Y

X
)− 2(

Z

X
)]W − (

Y

X
)(

Z

X
)Z}

= −x3+ 2x2z − xz2 + 72x2w − 17x yw+ y2w − 84xzw+ 10yzw + 24z2w;

Z2r ′′′(
Y

Z
,

X

Z
) = Z2{[(

Y

Z
)(

X

Z
) + (

X

Z
)− 2(

Y

Z
)]W − Z}

= −x3+ 3x2z − 3xz2 + z3 − 18x2w + 2x yw+ 18xzw− yzw − 4z2w,

X r ′′1 (
Y

X
,

Z

X
) = X {[(s+ t − 2t(

Y

X
)]W − t(

Z

X
)Z}

= −8xws+ yws+ 4zws− x2t + 2xzt − z2 t − 10xwt + ywt + 4zwt.

Example 2. Consider the non-singular rational degree 7 space curve given by

F(s, t) = (s7, s6 t, st6, t7).

Compute a µ-basis for F(s, t): p = ys− x t, q = ws− zt, r = zs5 − y t5.

The minimal generators for the Rees algebra associated to this curve consist of p,q, r, the

following quadric and sextic implicit equations of the space curve:

Sylvs,t(p,q) = xw− yz;

z5r ′i (
w

z
,

y

x
) = z5[z − y(

w

z
)5] = z6 − yw5;

xz4r ′i (
w

z
,

y

x
) = xz4[z − y(

w

z
)4(

y

x
)] = xz5 − y2w4;

x2z3r ′i (
w

z
,

y

x
) = x2z3[z − y(

w

z
)3(

y

x
)2] = x2z4 − y3w3;

x3z2r ′i (
w

z
,

y

x
) = x3z2[z − y(

w

z
)2(

y

x
)3] = x3z3 − y4w2;

x4zr ′i (
w

z
,

y

x
) = x4z[z − y(

w

z
)(

y

x
)4] = x4z2 − y5w;

x5r ′i (
w

z
,

y

x
) = x5[z − y(

y

x
)5] = x5z − y6,

together with
d(d − 3)

2
=

7(4)

2
= 14 moving planes of degree (d − 2− i, i + 1),

x jz i− j r ′j,i− j(
w

z
,

y

x
), ∀i = 1, . . . , d − 3= 4, 0≤ j ≤ i, i.e.,

z[zs4 − y t4(
w

z
)] = z2s4 − ywt4, x[zs4 − y t4(

y

x
)] = xzs4 − y2 t4,

z2[zs3 − y t3(
w

z
)2] = z3s3 − yw2 t3, xz[zs3 − y t3(

y

x
)(

w

z
)] = xz2s3 − y2wt3,

x2[zs3 − y t3(
y

x
)2] = x2zs3 − y3 t3, z3[zs2 − y t2(

w

z
)3] = z4s2 − yw3 t2,

xz2[zs2 − y t2(
y

x
)(

w

z
)2] = xz3s2 − y2w2 t2, x2z[zs2 − y t2(

y

x
)2(

w

z
)] = x2z2s2 − y3wt2,



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 620

x3[zs2 − y t2(
y

x
)3] = x3zs2 − y4 t2, z4[zs− y t(

w

z
)4] = z5s− yw4 t,

xz3[zs− y t(
y

x
)(

w

z
)3] = xz4s− y2w3 t, x2z2[zs− y t(

y

x
)2(

w

z
)2] = x2z3s− y3w2 t,

x3z[zs − y t(
y

x
)3(

w

z
)] = x3z2s− y4wt, x4[zs− y t(

y

x
)4] = x4zs− y5 t.

4. Rational Quartic Space Curves

In this section, we are going to discuss some of the geometry behind the generators for the

Rees algebra by studying rational quartic space curves. First, we would like to find implicit

equations for rational quartic space curves. We need to consider both singular and non-

singular curves. We begin with some generic results.

Lemma 4. If a rational quartic space curve C is singular, then the curve C is a complete inter-

section of two quadric surfaces.

If a rational quartic space curve C is non-singular, then the curve C is the projection of a

rational normal quartic curve S(4) ∈ P4 from a point Q /∈ S(4).

Proof. It is known [14, Page 353] that the geometric genus of every rational quartic space

curve C is zero, although the arithmetic genus of C may be either zero or one.

If C is singular, then the arithmetic genus of C is one, and C is a complete intersection

of two quadric surfaces [13, Chapter 1, page 44].

IfC is non-singular, then the arithmetic genus ofC is zero, andC is contained in a unique

non-singular quadric surface. Hence, by Theorem 6 [27], a rational non-singular quartic space

curve is the projection of a rational normal curve S(4) ∈ P4 from a point Q /∈ S(4).

By Lemma 1, we know that a rational quartic space curve is contained in the quadric

surface given Sylvs,t(p,q) = 0. Next we provide methods for finding implicit equations for

both singular and non-singular rational quartic space curves using moving planes and µ-bases.

4.1. Implicit Equations of Singular Rational Quartic Space Curves

By Proposition 2, for a singular rational quartic space curve C with a µ-basis p,q, r, the

only singular point P is the double point at the intersection of the axes of p and q.

Now suppose that P is a singular point on a rational quartic space curve C . If we take two

distinct points F(s1, t1), F(s2, t2) different from the singular point P on the space curveC , then

by Theorem 3.3 [26] there are two moving planes L1,L2 of degree one that follow the space

curve C with axes PF(s1, t1) and PF(s2, t2). Moreover we can easily choose F(s1, t1),F(s2, t2)

so that these axes are distinct. Since the axes of L1,L2 are distinct, the moving planes L1,L2

are linearly independent. Therefore, L1,L2, r form another µ-basis for the curve C .

Hence, without loss of generality, we can assume that the axes of p and q intersect the

space curve C at two distinct points F(s1, t1) and F(s2, t2) other than the singular point P. In

this case, the plane that contains the axes of p and q has the implicit equation

a(X) = [F(s1, t1),F(s2, t2), P] ·X= 0, (10)



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 621

where X= (x , y, z, w) and [·] denotes the outer product.

Next we are going to find two quadric surfaces that contain the singular rational quartic

space curve C .

To compute the implicit equations of a singular rational quartic space curve, we write the

µ-basis in the form:

p = p1s+ p0 t, q = q1s+ q0t, r = r2s2 + r1st + r0 t2,

where p1, p0,q1,q0, r2, r1, r0 are homogeneous polynomials of degree 1 in x , y, z, w. Let

M(p,q, r) be the 3 × 3 coefficient matrix of the moving planes (t1s − s1 t)p, (t2s − s2 t)q, r.

Since

(t1s− s1 t)p = t1p1s2 + (t1p0 − s1p1)st − s1p0 t2,

(t2s− s2 t)q = t2q1s2 + (t2q0− s2q1)st − s2q0t2,

it follows that

M(p,q, r) =







t1p1 t1p0 − s1p1 −s1p0

t2q1 t2q0− s2q1 −s2q0

r2 r1 r0






. (11)

Now we quote some results concerning the geometric construction of the implicit equations

of the singular quartic space curve. Detailed proofs can be found in [17, Theorem 4.9 and

Theorem 4.11].

Theorem 5. Let C be a singular rational quartic space curve, and let M = M(p,q, r) be the

matrix in Equation (11) constructed from a µ-basis for the curve. Then

det(M) = ah,

where a(x , y, z, w) = 0 is the implicit equation of the plane that contains the axes of p and q,

and h(x , y, z, w) = 0 is a quadric surface that contains the singular rational quartic space curve

C .

Theorem 6. Let C be a singular rational quartic space curve, and let M = M(p,q, r) be the

matrix in Equation (11) constructed from a µ-basis p,q, r for the curve C . Suppose nthe plane

that contains the axes of p,q has implicit equation a = 0. Then the two quadric surfaces f =

Sylvs,t(p,q) = 0 and g =
det(M)

a
= 0 form set-theoretic complete intersection generators for the

curve C .

Next we will use a very simple example to illustrate our method for finding both the

singular point and the implicit equations for a singular rational quartic space curve.

Example 3. Let the singular rational quartic space curve C be given as the image of the param-

eterization:

(x , y, z, w) = (s4, s3 t, s2 t2, t4).



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 622

Compute a µ-basis using the algorithm in [24]:

p = (y − z)s+ (y − x)t, q = zs− y t, r = ws2 − zt2.

The axis of p is the line through the points (0,0,0,1) and (1,1,1,1), and the axis of q is the line

through the points (1,0,0,0) and (0,0,0,1). These axes intersect at the point (0,0,0,1)—the

only singular point, a point of order 2—which corresponds to the parameters (s0, t0) = (0,1).

The curve C intersects the axis of p at the point (1,1,1,1) with parameters (s1, t1) = (1,1), and

intersects the axis of q at the point (1,0,0,0) with parameters (s2, t2) = (1,0). Now

f = Sylvs,t(p,q) = det

�

y − z y − x

z −y

�

= xz− y2,

g = det(M(p,q, r)) = det







t1p1 t1p0 − s1p1 −s1p0

t2q1 t2q0− s2q1 −s2q0

r2 r1 r0





 = det







y − z z − x x − y

0 −z y

w 0 −z







= z2 y − x yw− z3 + xzw.

The implicit equation of the plane containing the axes of p(s, t) and q(s, t) is:

a =

�

�

�

�

�

�

�

�

x y z w

1 1 1 1

1 0 0 0

0 0 0 1

�

�

�

�

�

�

�

�

= y − z = 0.

Therefore the other quadric surface that contains the quartic curve is

h=
g

a
= z2 − xw.

Hence the implicit equations of the singular rational quartic curve C are (see Figure 1)

xz − y2 = 0, z2 − xw = 0.

4.2. Implicit Equations of Non-Singular Rational Quartic Space Curves

Next we consider the case where the rational quartic space curveC given by the parametriza-

tion f0, f1, f2, f3 is smooth. By Lemma 4, we know that the curve C is the image of the

projection of the rational normal scroll S(4) from a point P /∈ S(4)⊂ P4 to a hyperplane in P4.

Lemma 5. The implicit equations of a non-singular rational quartic space curve C are given

by 4 equations F1 = 0, F2 = 0, F3 = 0, F4 = 0, one quadric surface and three cubic surfaces.

Moreover, the quadric surface is given by F1 = Sylvs,t(p,q) = 0, where p,q are the degree 1

elements of a µ-basis for C .



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 623

Figure 1: Set-theoreti
 
omplete interse
tion of a singular rational quarti
 spa
e 
urve.
Proof. Let P = (a, b, c, d , 1) ∈ P4 be a point not contained in the rational normal curve

S(4) given by the parametric representation

x = s4, y = s3 t, z = s2 t2, w = st3, u = t4.

Without loss of generality, let the non-singular rational quartic space curve C be the image

of the projection of the rational normal curve S(4) from the point P to the hyperplane u = 0.

Observe that the parametrization of the image can be described as F(s, t) = (s4 − at4, s3 t −
bt4, s2 t2 − ct4, st3 − d t4). Therefore the implicit equations of C are the generators of the

following ideal:

〈x − (s4 − at4), y − (s3t − bt4), z − (s2t2 − ct4), w − (st3 − d t4)〉
⋂

K(a, b, c, d)[x , y, z, w].

With the aid of a computer algebra system, we compute four polynomial generators: F1, F2,

F3, F4; with one quadric F1 and three cubics F2, F3, F4–see below. Moreover, F1 = Sylvs,t(p,q),

since by Lemma 4 any non-singular quartic space curve is contained in exactly one quadric.

F1 =(−c + d2)xz+ (b− cd)xw+ (c − d2)y2 + (−b+ cd)yz + (−a+ 2bd − c2)yw+

(a− bd)z2 + (−ad + bc)zw + (ac − b2)w2;

F2 =(c − d2)y2w + (−c+ d2)yz2 + (−b+ cd)yzw + (2bd − 2c2)yw2 + (b− d3)z3+

(−3bd + 3cd2)z2w + (3bc− 3c2d)zw2 + (−b2+ c3)w3;

F3 =(−c2 + 2cd2 − d4)x yw+ (b2 − 3bcd + bd3+ c3)xw2 + (c2 − 2cd2 + d4)y2z+

(bc − bd2− c2d + cd3)y2w + (−bc+ bd2+ cd3 − d5)yz2 + (−b2 + 4bcd − 2bd3−

3c2d2+ 2cd4)yzw + (−ab+ ad3 + 2b2d − bc2 − 2bcd2 + 2c3d − c2d3)yw2+

(ab− ad3− b2d + bd4)z2w + (−abd − ac2 + 2acd2 + b2d2+ bc2d − 2bcd3)zw2+

(2abc − abd2− ac2d − b3 + b2cd − bc3 + bc2d2)w3;

F4 =(−c2 + 2cd2 − d4)x2w + (c2 − 2cd2+ d4)x yz + (bc − bd2 − c2d + cd3)x yw+

(−bc + bd2+ cd3 − d5)xz2+ (bcd − bd3+ c3 − 3c2d2+ 2cd4)xzw+ (ab−



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 624

3acd + 2ad3− bcd2 + 2c3d − c2d3)xw2+ (−ab+ 3acd − 2ad3− bc2 + 2bcd2−

c3d)yzw + (−a2+ 2abd − 2bc2d + c4)yw2 + (a2− abd − ac2 − acd2 + ad4+

b2c − b2d2+ bc2d)z2w + (−a2d − abc + 2abd2+ 3ac2d − 2acd3−

bc3)zw2 + (2a2c − a2d2 − ab2 − 2ac3 + ac2d2+ b2c2)w3.

Lemma 5 is a theoretical result. Our next goal is to find simple explicit expressions for

these four implicit equations. To proceed, we first dehomogenize the µ-basis elements by

setting t = 1, so that

p = p1s+ p0, q = q1s+ q0, r = r2s2 + r1s+ r0,

where deg(p1) = deg(p0) = deg(q1) = deg(q0) = deg(r2) = deg(r1) = deg(r1) = 1.

If we take the resultants with respect to the variable s, then we generate the following

three expressions:

Res(p,q) = det

�

p1 p0

q1 q0

�

= Sylvs,t(p,q) = p1q0 − p0q1,

Res(p, r) = det







p1 0 r2

p0 p1 r1

0 p0 r0





 = r2p2
0 − r1p1p0 + r0p2

1,

Res(q, r) = det







q1 0 r2

q0 q1 r1

0 q0 r0





 = r2q2
0 − r1q1q0 + r0q2

1,

where deg(Res(p,q)) = 2, deg(Res(p, r)) = 3, and deg(Res(q, r)) = 3 in x , y, z, w.

Remark 4. If p,q, r are a µ-basis for the space curve C , then C is contained in the three surfaces

Res(p,q) = 0, Res(p, r) = 0 and Res(q, r) = 0 because each element of the µ-basis follows the

curve.

Lemma 6.

gcd(Res(p,q),Res(p, r)) = gcd(Res(p,q),Res(q, r)) = gcd(Res(p, r),Res(q, r)) = 1.

Proof. Suppose that

gcd(Res(p,q),Res(p, r)) = g 6= 1.

Then for any point A satisfying g(A) = 0, we have

deg(gcd(p(s, t,A),q(s, t,A))) ≥ 1, and deg(gcd(p(s, t,A), r(s, t,A))) ≥ 1.

Since deg(p) = deg(q) = 1 and deg(r) = 2 in the variables s, t, we have

deg(gcd(p(s, t,A),q(s, t,A), r(s, t,A))) ≥ 1.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 625

Therefore by Proposition 1 the point A is on the rational space curve C . Hence the surface

g = 0 is contained in the space curve C . This is impossible. Therefore,

gcd(Res(p,q),Res(p, r)) = 1. A similar argument applies to show that

gcd(Res(p,q), Res(q, r)) = 1.

Now suppose that gcd(Res(p, r),Res(q, r)) = g 6= 1.

If deg(g) = 3, then Res(p, r) = kRes(q, r) for some nonzero constant k. Hence since the

axis of p is contained in the cubic surface Res(p, r) = 0, the axis of p is also contained in

the surface Res(q, r) = 0. Thus for a point A on the axis of p, we have p(s, t,A) = 0 and

deg(gcd(q(s, t,A), r(s, t,A))) ≥ 1. Hence

deg(gcd(p(s, t,A),q(s, t,A), r(s, t,A))) ≥ 1,

so the point A is on the curve C . Therefore, the axis line of p is contained in the rational space

curve C . This is impossible.

If deg(g) = 2, then since a rational non-singular quartic curve can be contained in only

one quadric surface, we would have g = Res(p,q), again contradicting the fact that

gcd(Res(p,q),Res(p, r)) = 1.

If deg(g) = 1, then since C is a space curve contained in the cubic surface Res(p, r) = 0,

we would have C is contained in the quadric surface
Res(p,r)

g
= 0. Again since the non-

singular quartic space curve C can be contained in only one quadric surface, we would have
Res(p,r)

g
= Res(p,q), again contradicting the fact that gcd(Res(p,q),Res(p, r)) = 1. Therefore,

gcd(Res(p, r),Res(q, r)) = 1.

Now let N(p,q, r) be the 3× 3 coefficient matrix of the moving planes sp, tq, r. Since

sp = p1s2 + p0st, tq = q1st + q0 t2, r = r2s2 + r1st + r0t2

it follows that

N(p,q, r) =







p1 0 r2

p0 q1 r1

0 q0 r0





 , and det(N) = r2p0q0 − r1p1q0 + r0p1q1. (12)

Lemma 7. det(N(p,q, r)) is not identically zero.

Proof. If det(N(p,q, r)) ≡ 0, then the rows of the matrix N are linearly dependent over

the ring K[x , y, z, w]. Without loss of generality, by Lemma 3 we can assume that p1 =

y, p0 = −x ,q1 = w,q0 = −z. A dependence relation among the three rows of N(p,q, r) would

generate three linear equations

a y − bx = 0

bw − cz = 0

ar2 + br1+ cr0 = 0

with homogeneous polynomials a(x , y, z, w), b(x , y, z, w), c(x , y, z, w) of the same degree in

x , y, z, w at least one of which is not zero. From the first equation, we find that a = αx , b =



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 626

αy. The second equation then shows that α = βz, c = β yw. Substituting these results into

the third equation gives xzr2 + yzr1 + ywr0 = 0. In other words, (r2, r1, r0) is a syzygy of

(xz, yz, yw). But the syzygy module for (xz, yz, yw) is generated by (y,−x , 0) and (0, w,−z).

Therefore, r2 = my, r1 = −mx + nw, r0 = −nz where m, n are necessarily constants because

the ri are linear in x , y, z, w. Hence, column three of the matrix N is a K-linear combination

of the first two columns. Thus, the moving plane r is a K-linear combination of the moving

planes sp and tq. But this is impossible because p,q, r are linearly independent over the ring

K[s, t].

The surface det(N(p,q, r)) = 0 is a cubic surface containing the non-singular rational

quartic space curve C because the moving planes p,q, r follow the curve C . Moreover, we

have the following result.

Theorem 7. Res(p,q) = 0, Res(p, r) = 0, Res(q, r) = 0, det(N(p,q, r)) = 0 are the implicit

equations of the non-singular rational quartic space curve with µ-basis p,q, r.

Proof. Since p,q, r are a µ-basis, the space curve C is contained in the surface

det(N(p,q, r)) = 0. By Remark 4, the space curve C is also contained in the surfaces

Res(p,q) = 0, Res(p, r) = 0, and Res(q, r) = 0. Hence the polynomials Res(p,q), Res(p, r),

Res(q, r), and det(N(p,q, r)) are contained in the ideal of the quartic space curve C , which is

generated by the four polynomials F1, F2, F3, F4 in Lemma 5.

To simplify our notation, let N = N(p,q, r). Now we claim that the cubic surface det(N) =

0 is irreducible. Otherwise, the cubic det(N) would have Res(p,q) as a factor, since a non-

singular rational quartic space curve C is contained in exactly one quadric surface. But this

is impossible, since there exists at least one point which lies on the surface Res(p,q) = 0, but

not on the surface det(N) = 0. To see that such a point exists, recall that by Lemma 3 there

is a linear transformation on the µ-basis elements p,q and a projective change of coordinates

so that p = ys − x t, q = ws− zt, r = r2s2 + r1st + r0 t2, and det(N) = r2 xz + r1 yz + r0 yw.

Now observe that the surface Res(p,q) = xw − yz = 0 contains the line x = z = 0. But there

must be a point P = (0, a, 0, b) on the line x = z = 0 such that det(N)(P) = r0(P)ab 6= 0. If

not, then r0(P) = 0, so deg(gcd(p · P, q · P, r · P)) = deg(gcd(as, bs, r2s2+ r1st)) = 1. Hence by

Proposition 1, C must contain a line, contradicting the assumption that C is a space curve.

Moreover, the three cubic surfaces det(N) = 0, Res(p, r) = 0 and Res(q, r) = 0 are distinct.

Indeed, since both p and q are axial moving planes, the axis of p is contained in the surfaces

Res(p, r) = 0 and det(N) = 0, but is not contained in the surface Res(q, r) = 0; and similarly,

the axis of q is contained in the surfaces Res(q, r) = 0 and det(N) = 0, but is not contained

in the surface Res(p, r) = 0. Indeed a point X is on Res(q, r) = 0 if and only if there is an

(s, t) such that X is on the planes q(s, t) = 0, r(s, t) = 0. If X is also on the axis of p, then the

point X would also be on p(s, t) = 0, and thus by Proposition 1 on the curve C . Therefore

the entire axis of p would be on C , which is absurd. Hence the three surfaces det(N) = 0,

Res(p, r) = 0 and Res(q, r) = 0 are different from each other.

Furthermore, the three cubic surfaces det(N) = 0, Res(p, r) = 0 and Res(q, r) = 0 are

linearly independent. For suppose that there exist some nonzero constants a, b, c such that

aRes(p, r) + bRes(q, r)+ c det(N)≡ 0.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 627

Then

〈Res(p, r), det(N)〉= 〈Res(p, r),Res(q, r)〉,

so

V(〈Res(p, r), det(N)〉) = V(Res(p, r),Res(q, r)〉).

Hence the axis of p is contained in the surface Res(q, r) = 0. Contradiction. Therefore, the

three cubic surfaces det(N) = 0, Res(p, r) = 0 and Res(q, r) = 0 are linearly independent.

Thus, by Lemma 5 {Res(p,q), Res(p, r), Res(q, r), det(N)} must be a set of generators

for the ideal of the non-singular rational quartic space curve C . Therefore, Res(p,q) =

0, Res(p, r) = 0, Res(q, r) = 0, det(N) = 0 are the implicit equations of the non-singular

rational quartic space curve.

We illustrate our method for finding the implicit equations of a non-singular rational quar-

tic space curve with the following simple example.

Example 4. Let the non-singular rational quartic space curve C be given as the image of the

parameterization:

(x , y, z, w) = (s4, s2 t(s+ t), st2(s− t), t4).

Compute a µ-basis using the algorithm in [24]

p = 2t x + (−2s+ t)y + sz, q = t x − s y + (s+ t)z + sw, r = (−t2 − st)x + s2 y.

Then

Res(p,q) = y2 − xz− 3yz + z2 − 2xw− yw,

Res(p, r) = z3 − xzw + 2yzw − z2w − 2xw2+ yw2,

Res(q, r) = yz2 − x yw+ 2xzw + 3yzw − z2w + yw2,

det(N(p,q, r)) = xz2 − x2w + 2x yw− 3xzw − 3yzw + z2w − xw2 − yw2

are the implicit equations of the curve C (see Figure 2).

Figure 2: Set-theoreti
 generators of a non-singular rational quarti
 spa
e 
urve.



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 628

4.3. Generators for the Rees Algebra Associated to Rational Quartic Space Curves

Now we can give a minimal set of generators for the Rees algebra associated to a rational

quartic space curve using only the µ-basis of the curve.

Theorem 8. Let p,q, r be a µ-basis for a rational quartic space curve. Then:

1. A minimal set of generators for the kernel K of the Rees algebra for a singular rational

quartic space curve is given by

p, q, r, Sylvs,t(p,q),
det(M(p,q, r))

a(x , y, z, w)
,

where a and M are defined in Equations (10) and (11).

2. A minimal set of generators for the kernel K of the Rees algebra for a non-singular rational

quartic space curve is given by

p, q, r, Sylvs,t(p, r), Sylvs,t(q, r), Res(p,q), Res(p, r), Res(q, r), det(N(p,q, r)),

where N(p,q, r) is defined in Equation (12).

Proof. We will prove the claim by comparing the generators listed above against the gen-

erators described in Theorems 1 and 3. First, we note that the µ-basis elements p,q, r and

Sylvs,t(p,q) = Res(p,q) are among the generators in both theorems. We will focus therefore

on the other generators.

For singular quartic space curves, recall that by Lemma 2 there is a linear transformation

on the µ-basis elements p,q, and a projective change of coordinates so that p = ys− x t, q =

zs− y t, and the singular point is located at (0,0,0,1). Now the axial plane determined by the

axes of p and q is defined by a(x , y, z, w) = y = 0.

In addition, notice that p(1,0) = y and q(0,1) = −y. Therefore, since the µ-basis ele-

ments p(s, t) and q(s, t) follow the curve F(s, t), the quartic space curve F(s, t) intersects the

axial plane y = 0 at the two points F(0,1) and F(1,0). Hence st must be a factor of y, so

y = stβ , where β is a homogenous form of degree 2 in s, t. Moreover, since the two µ-basis

elements p = ys − x t, q = zs − y t follow the curve F(s, t), we conclude that on the curve

F(s, t):

x = s2β , y = stβ , z = t2β , where β is a homogeneous form in s, t of degree 2. (13)

In fact, the roots of β(s, t) = 0 are the two parameters corresponding to the singular point

(0,0,0,1) on the curve F(s, t). We will study the generator det(M(p,q, r))/a(x , y, z, w) by

investigating the roots of the polynomial β(s, t) = 0 in the following three cases.

Case 1: Neither (0,1) nor (1,0) is a root of β(s, t) = 0. Then F(0,1) and F(1,0) are both non-

singular points on the curve F(s, t). Setting (s1, t1) = (0,1) and (s2, t2) = (1,0) in

Equation (11) yields:

det(M(p,q, r)) = det







y −x 0

0 −z y

r2 r1 r0






= −r2 x y − r1 y2 − r0 yz;



J. Hoffman, H. Wang, X. Jia, R. Goldman / Eur. J. Pure Appl. Math, 3 (2010), 602-632 629

det(M(p,q, r))

a(x , y, z, w)
= −[r2 x + r1 y + r0z] = −x[r2+ r1(

y

x
) + r0(

z

x
)] = −x r ′1(

y

x
,

z

x
).

Thus, the above generators are the same as the generators described in Theorem 1.

Case 2: Either (0,1) or (1,0) is a root of β(s, t) = 0, but β(1,1) 6= 0 and β(1,−1) 6= 0. Then by

Equation (13) F(s1, t1) = F(1,−1) = (1,−1,1,∗) and F(s2, t2) = F(1,1) = (1,1,1,∗) are

two distinct non-singular points on the curve F(s, t). Therefore, we may choose a new

µ-basis

p+ q = (y + z)s− (x + y)t, p− q = (y − z)s− (x − y)t, r.

Now the axial moving plane is a(x , y, z, w) = det











x y z w

0 0 0 1

1 1 1 ∗
1 −1 1 ∗











= 2(x − z) = 0. In

this case, Equation (11) again yields:

det(M(p,q, r)) = det







−(y + z) x − z x + y

y − z −x + z x − y

r2 r1 r0





 = 2(x − z)[r2 x + r1 y + r0z];

det(M(p,q, r)

a(x , y, z, w)
= −[r2 x + r1 y + r0z] = −x[r2+ r1(

y

x
) + r0(

z

x
)] = −x r ′1(

y

x
,

z

x
).

Thus, the above generators are the same as the generators described in Theorem 1.

Case 3: Either (0,1) or (1,0) and either (1,1) or (1,−1) is a root of β(s, t) = 0. Without loss

of generality, assume that β(0,1) 6= 0 and β(1,−1) 6= 0. Then F(1,0) = F(1,1) =

(0,0,0,1) is the singular point, and by Equation (13) F(s1, t1) = F(0,1) = (0,0,1,∗)
and F(s2, t2) = F(1,−1) = (1,−1,1,∗) are two distinct non-singular point on the curve

F(s, t). Therefore, we may choose a new µ-basis

p = ys− x t, p+ q = (y + z)s− (x + y)t, r.

Now the axial moving plane is a(x , y, z, w) = det











x y z w

0 0 0 1

0 0 1 ∗
1 −1 1 ∗











= x + y = 0. In

this case, Equation (11) again gives:

det(M(p,q, r)) = det







y −x 0

−(y + z) x − z x + y

r2 r1 r0






= −(x + y)(r2 x + r1 y + r0z);

det(M(p,q, r))

a(x , y, z, w)
= −[r2 x + r1 y + r0z] = −x[r2+ r1(

y

x
) + r0(

z

x
)] = −x r ′1(

y

x
,

z

x
).

Thus, the above generators are the same as the generators described in Theorem 1.



REFERENCES 630

For non-singular quartic space curves, recall that by Lemma 3 there is a linear trans-

formation on the µ-basis elements p,q, and a projective change of coordinates so that p =

ys− x t, q = ws− zt. Then

Res(p, r) = r2 x2+ r1 x y + r0 y2 = x2[r2 + r1(
y

x
) + r0(

y

x
)2] = x2r ′2(

y

x
,

w

z
),

Res(q, r) = r2z2 + r1zw + r0w2 = z2[r2 + r1(
w

z
) + r0(

w

z
)2] = z2r ′0(

y

x
,
w

z
),

det(N) = r2 xz+ r1 xw+ r0 yw = xz[r2 + r1(
w

z
) + r0(

w

z
)(

y

x
)] = xzr ′1(

y

x
,
w

z
),

Sylvs,t(p, r) = r2 xs+ r1 x t + r0 y t = x[r2s+ r1t + r0t(
y

x
)] = x r ′1,0(

y

x
,

w

z
),

Sylvs,t(q, r) = r2zs+ r1zt + r0wt = z[r2s+ r1 t + r0 t(
w

z
)] = zr ′0,1(

y

x
,
w

z
).

Thus, the generators listed in the statement of this theorem are exactly the same as the gen-

erators listed in the statement of Theorem 3.

ACKNOWLEDGEMENTS We would like to thank David Cox for his helpful suggestions.

This work was partially supported by NSF grant CCR-020331, and by the NSF of China

(No.60873109), One Hundred Talent Project supported by CAS and the 111 Project (No.

B07033). The second author would like to thank the organizers of the Special Session on

Geometry of Varieties, Syzygies and Computations for giving her the opportunity to present

this paper during the first KMS-AMS joint meeting.

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